% LaTeX template, xeCJK usepackage and Unicode text need XeLaTeX to compile.
% MiKTeX package can be downloaded at miktex.org, WinEdt can be downloaded at http://www.winedt.com/.
% WinEdt 6 and higher provide XeLaTeX and Unicode support.

% LaTeX template version 1.0.3, last revised on 2014-11-14.

\documentclass[10pt]{article}



% *************************** installed packages *************************
% AMS packages
\usepackage{amsfonts}   % TeX fonts from the American Mathematical Society.
\usepackage{amsmath}    % AMS mathematical facilities for LaTeX.
\usepackage{amssymb}    % AMS symbols
\usepackage{amsthm}     % Provide proclamations environment.

% graphics packages
\usepackage{graphics}   % Standard LaTeX graphics.
\usepackage{tikz}       % TikZ and PGF package for graphics
\usetikzlibrary{matrix} % matrix library of TikZ package
\usetikzlibrary{trees}  % trees library of TikZ package

% support for foreign languages, esp. Chinese.
\usepackage{xeCJK}      % Support for CJK (Chinese, Japanese, Korean) documents in XeLaTeX.
\setCJKmainfont{SimSun} % MUST appear when xeCJK is loaded.
%\setCJKmainfont{DFKai-SB}          % 设置正文罗马族的CKJ字体，影响 \rmfamily 和 \textrm 的字体。此处设为“标楷体”。
%\setCJKmainfont{SimSun}            % 设置正文罗马族的CKJ字体，影响 \rmfamily 和 \textrm 的字体。此处设为“宋体”。
%\setCJKmonofont{MingLiU}           % 设置正文等宽族的CJK字体，影响 \ttfamily 和 \texttt 的字体。此处设为“细明体”。
%\renewcommand\abstractname{摘要}   % 重定义摘要名：abstract->摘要。
%\renewcommand\appendixname{附录}   % 重定义附录名：appendix->附录。
%\renewcommand\bibname{参考文献}    % 重定义参考文献名：bibliography->参考文献。
%\renewcommand\contentsname{目录}   % 重定义目录名：contents->目录。
%\renewcommand\refname{参考文献}    % 重定义参考文献名：references->参考文献。

% miscellaneous packages
\usepackage[toc, page]{appendix}   % Extra control of appendices.
\usepackage{caption}    % Customising captions in floating environments.
%\captionsetup[figure]{labelformat=empty} % redefines the caption setup of the figures environment in the beamer %class.
\usepackage{clrscode}   % Typesets pseudocode as in Introduction to Algorithms.
\usepackage{epsfig}
\usepackage{eurosym}    % Metafont and macros for Euro sign.
\usepackage{float}      % Improved interface for floating objects.
\usepackage{fontspec}   % Advanced font selection in XeLaTeX and LuaLaTeX.
\usepackage{xcolor}     % Driver-independent color extensions for LaTeX and pdfLaTeX.

% must-be-the-last packages
\usepackage[driverfallback=hypertex, pagebackref]{hyperref}   % Extensive support for hypertext in LaTeX; MUST be on the last \usepackage line in the preamble. [pagebackref] for page referencing; [backref] for section referencing.
% ********************** end of installed packages ***********************



% ************************** fullpage.sty ********************************
% This is FULLPAGE.STY by H.Partl, Version 2 as of 15 Dec 1988.
% Document Style Option to fill the paper just like Plain TeX.
\typeout{Style Option FULLPAGE Version 2 as of 15 Dec 1988}

\topmargin 0pt \advance \topmargin by -\headheight \advance
\topmargin by -\headsep

\textheight 8.9in

\oddsidemargin 0pt \evensidemargin \oddsidemargin \marginparwidth
0.5in

\textwidth 6.5in
% For users of A4 paper: The above values are suited for American 8.5x11in
% paper. If your output driver performs a conversion for A4 paper, keep
% those values. If your output driver conforms to the TeX standard (1in/1in),
% then you should add the following commands to center the text on A4 paper:

% \advance\hoffset by -3mm  % A4 is narrower.
% \advance\voffset by  8mm  % A4 is taller.
% ************************ end of fullpage.sty ***************************



% ************** Proclamations (theorem-like structures) *****************
% [section] option provides numbering within a section.
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}[section]
\newtheorem{definition}{Definition}[section]
\newtheorem{prop}{Proposition}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{remark}{Remark}
\newtheorem{example}{Example}
% ************************************************************************



% ************* Solutions use a modified proof environment ***************
\newenvironment{solution}
  {\begin{proof}[Solution]}
  {\end{proof}}
% ************************************************************************



% ************* Frequently used commands as shorthand ********************
\newcommand{\norm}{|\!|}
% ************************************************************************



\begin{document}
\title{ \Huge{\it Stochastic Calculus for Finance II: Continuous-Time Models} \\ {Solution of Exercise Problems} }
\author{Yan Zeng}
\date{Version 1.0.8, last revised on 2015-03-13.}

\maketitle

\begin{abstract}
This is a solution manual for Shreve \cite{Shreve04b}. If you find any typos/errors or have any comments, please email me at zypublic@hotmail.edu. This version skips Exercise 7.1, 7.2, 7.5--7.9.
\end{abstract}

\tableofcontents

\newpage

\section{General Probability Theory}

$\bigstar$ {\bf Comments}:

\medskip

Example 1.1.4 of the textbook illustrates the paradigm of extending probability measure from a $\sigma$-algebra consisting of finitely many elements to a $\sigma$-algebra consisting of infinitely many elements. This procedure is typical of constructing probability measures and its validity is justified by Carath\'{e}odary's Extension Theorem (see, for example, Durrett\cite[page~402]{Durrett10}).

\bigskip

\noindent  $\blacktriangleright$  {\bf Exercise 1.1.} using the properties of Definition 1.1.2 for a probability measure ${\mathbb P}$, show the following.

\smallskip

\noindent (i) If $A \in {\cal F}$, $B \in {\cal F}$, and $A \subset B$, then ${\mathbb P}(A) \le {\mathbb P}(B)$.

\begin{proof}
${\mathbb P}(B) = {\mathbb P}((B-A)\cup A) = {\mathbb P}(B-A)+P(A)\ge {\mathbb P}(A)$.
\end{proof}

\noindent (ii) If $A \in {\cal F}$ and $\{A_n\}_{n=1}^{\infty}$ is a sequence of sets in ${\cal F}$ with $\lim_{n\to\infty} {\mathbb P}(A_n) = 0$ and $A \subset A_n$ for every $n$, then ${\mathbb P}(A)=0$. (This property was used implicitly in Example 1.1.4 when we argued that the sequence of all heads, and indeed any particular sequence, must have probability zero.)

\begin{proof}
According to (i), ${\mathbb P}(A)\le {\mathbb P}(A_n)$, which implies ${\mathbb P}(A)\le
\lim_{n\to\infty}{\mathbb P}(A_n)=0$. So $0\le {\mathbb P}(A)\le 0$. This means
${\mathbb P}(A)=0$.
\end{proof}

\medskip

\noindent  $\blacktriangleright$  {\bf Exercise 1.2.} The infinite coin-toss space $\Omega_{\infty}$ of Example 1.1.4 is {\it uncountably infinite}. In other words, we cannot list all its elements in a sequence. To see that this is impossible, suppose there were such a sequential list of all elements of $\Omega_{\infty}$:
\begin{eqnarray*}
\omega^{(1)} &=& \omega_1^{(1)}\omega_2^{(1)}\omega_3^{(1)}\omega_4^{(1)} \cdots, \\
\omega^{(2)} &=& \omega_1^{(2)}\omega_2^{(2)}\omega_3^{(2)}\omega_4^{(2)} \cdots, \\
\omega^{(3)} &=& \omega_1^{(3)}\omega_2^{(3)}\omega_3^{(3)}\omega_4^{(3)} \cdots, \\
             &\vdots&
\end{eqnarray*}
An element that does not appear in this list is the sequence whose first component is $H$ is $\omega_1^{(1)}$ is $T$ and is $T$ if $\omega_1^{(1)}$ is $H$, whose second component is $H$ if $\omega_2^{(2)}$ is $T$ and is $T$ if $\omega_2^{(2)}$ is $H$, whose third component is $H$ if $\omega_3^{(3)}$ is $T$ and is $T$ if $\omega_3^{(3)}$ is $H$, etc. Thus, the list does not include every element of $\Omega_{\infty}$.

Now consider the set of sequences of coin tosses in which the outcome on each even-numbered toss matches the outcome of the toss preceding it, i.e.,
\[
A = \{\omega = \omega_1\omega_2\omega_3\omega_4\omega_5\cdots; \omega_1=\omega_2,\omega_3=\omega_4,\cdots\}.
\]

\smallskip

\noindent (i) Show that $A$ is uncountably infinite.

\begin{proof}
We define a mapping $\phi$ from $A$
to $\Omega_{\infty}$ as follows:
$\phi(\omega_1\omega_2\cdots)=\omega_1\omega_3\omega_5\cdots$. Then
$\phi$ is one-to-one and onto. So the cardinality of $A$ is the same
as that of $\Omega_{\infty}$, which means in particular that $A$ is
uncountably infinite.
\end{proof}

\noindent (ii) Show that, when $0<p<1$, we have ${\mathbb P}(A) = 0$.

\begin{proof} Let
$A_n=\{\omega=\omega_1\omega_2\cdots:\omega_1=\omega_2,\cdots,\omega_{2n-1}=\omega_{2n}\}$.
Then $A_n\downarrow A$ as $n\to\infty$. So
\[
{\mathbb P}(A)=\lim_{n\to\infty}{\mathbb P}(A_n)=\lim_{n\to\infty}[{\mathbb P}(\omega_1=\omega_2)\cdots
{\mathbb P}(\omega_{2n-1}=\omega_{2n})]=\lim_{n\to\infty}[p^2+(1-p)^2]^n.
\]
Since $p^2+(1-p)^2 \le \max\{p, 1-p\}[p+(1-p)]<1$ for $0<p<1$,
 we have $\lim_{n\to\infty}(p^2+(1-p)^2)^n=0$. This implies ${\mathbb P}(A)=0$.
\end{proof}

\medskip

\noindent  $\blacktriangleright$  {\bf Exercise 1.3.} Consider the set function ${\mathbb P}$ defined for every subset of $[0,1]$ by the formula that ${\mathbb P}(A)=0$ if $A$ is a finite set and ${\mathbb P}(A) = \infty$ if $A$ is an infinite set. Show that ${\mathbb P}$ satisfies (1.1.3)-(1.1.5), but ${\mathbb P}$ does not have the countable additivity property (1.1.2). We see then that the finite additivity property (1.1.5) does not imply the countable additivity property (1.1.2).

\begin{proof} Clearly ${\mathbb P}(\emptyset)=0$. For any $A$
and $B$, if both of them are finite, then $A\cup B$ is also finite.
So ${\mathbb P}(A\cup B)=0={\mathbb P}(A)+{\mathbb P}(B)$. If at least one of them is infinite,
then $A\cup B$ is also infinite. So ${\mathbb P}(A\cup B)=\infty={\mathbb P}(A)+{\mathbb P}(B)$.
Similarly, we can prove ${\mathbb P}(\cup_{n=1}^NA_n)=\sum_{n=1}^N{\mathbb P}(A_n)$,
even if $A_n$'s are not disjoint.

To see countable additivity property doesn't hold for ${\mathbb P}$, let
$A_n=\{\frac{1}{n}\}$. Then $A=\cup_{n=1}^{\infty}A_n$ is an
infinite set and therefore ${\mathbb P}(A)=\infty$. However, ${\mathbb P}(A_n)=0$ for
each $n$. So ${\mathbb P}(A)\ne \sum_{n=1}^{\infty}{\mathbb P}(A_n)$.
\end{proof}

\medskip

\noindent  $\blacktriangleright$  {\bf Exercise 1.4.}

\smallskip

\noindent (i) Construct a standard normal random variable $Z$ on the probability space $(\Omega_{\infty}, {\cal F}_{\infty}, {\mathbb P})$ of Example 1.1.4 under the assumption that the probability for head is $p=\frac{1}{2}$. (Hint: Consider Examples 1.2.5 and 1.2.6)

\begin{solution} By Example 1.2.5, we can construct
a random variable $X$ on the coin-toss space, which is uniformly
distributed on $[0,1]$. For the strictly increasing and continuous
function
$N(x)=\int_{-\infty}^x\frac{1}{\sqrt{2\pi}}e^{-\frac{\xi^2}{2}}d\xi$,
 we let $Z=N^{-1}(X)$. Then ${\mathbb P}(Z\le a)={\mathbb P}(X\le N(a))=N(a)$ for any real number $a$, i.e. $Z$
 is a standard normal random variable on the coin-toss space
 $(\Omega_{\infty},{\cal F}_{\infty},{\mathbb P})$.
\end{solution}

\noindent (ii) Define a sequence of random variables $\{Z_n\}_{n=1}^{\infty}$ on $\Omega_{\infty}$ such that
\[
\lim_{n\to\infty} Z_n(\omega) = Z(\omega) \; \mbox{for every $\omega\in \Omega_{\infty}$}
\]
and, for each $n$, $Z_n$ depends only on the first $n$ coin tosses. (This gives us a procedure for approximating a standard normal random variable by random variables generated by a finite number of coin tosses, a useful algorithm for Monte Carlo simulation).

\begin{solution} Define
\[
X_n=\sum_{i=1}^n\frac{1}{2^i} 1_{\{\omega_i=H\}}.
\]
Then $X_n(\omega)\to X(\omega)$ for every $\omega\in
\Omega_{\infty}$ where $X$ is defined as in Example 1.2.5. So
$Z_n=N^{-1}(X_n)\to Z=N^{-1}(X)$ for every $\omega$. Clearly $Z_n$
depends only on the first $n$ coin tosses and $\{Z_n\}_{n\ge 1}$ is
the desired sequence.
\end{solution}

\medskip

\noindent  $\blacktriangleright$  {\bf Exercise 1.5.} When dealing with double Lebesgue integrals, just as with double Riemann integrals, the order of integration can be reversed. The only assumption required is that the function being integrated be either nonnegative or integrable. Here is an application of this fact.

Let $X$ be a nonnegative random variable with cumulative distribution function $F(x) = {\mathbb P}\{ X \le x\}$. Show that
\[
{\mathbb E}X = \int_0^{\infty} (1-F(x))dx
\]
by showing that
\[
\int_{\Omega} \int_0^{\infty} 1_{[0, X(\omega))}(x) dx d{\mathbb P}(\omega)
\]
is equal to both ${\mathbb E}X$ and $\int_0^{\infty}(1-F(x))dx$.

\begin{proof} First, by the information given by the
problem, we have
\[
\int_{\Omega}\int_0^{\infty}1_{[0,X(\omega))}(x)dxd{\mathbb P}(\omega)=\int_0^{\infty}\int_{\Omega}1_{[0,X(\omega))}(x)d{\mathbb P}(\omega)dx.
\]The left side of this equation equals to
\[
\int_{\Omega}\int_0^{X(\omega)}dxdP(\omega)=\int_{\Omega}X(\omega)dP(\omega)={\mathbb P} [X].
\]The right side of the equation equals to
\[
\int_0^{\infty}\int_{\Omega}1_{\{x<X(\omega)\}}d{\mathbb P}(\omega)dx=\int_0^{\infty}{\mathbb P}(x<X)dx=\int_0^{\infty}(1-F(x))dx.
\]So ${\mathbb E}[X]=\int_0^{\infty}(1-F(x))dx$.
\end{proof}

\medskip

\noindent  $\blacktriangleright$  {\bf Exercise 1.6.} Let $u$ be a fixed number in ${\mathbb R}$, and define the convex function $\varphi(x)=e^{ux}$ for all $x\in \mathbb R$. Let $X$ be a normal random variable with mean $\mu = {\mathbb E}X$ and standard deviation $\sigma = [{\mathbb E}(X-\mu)^2]^{\frac{1}{2}}$, i.e., with density
\[
f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}.
\]

\smallskip

\noindent (i) Verify that
\[
{\mathbb E} e^{uX} = e^{u\mu+\frac{1}{2}u^2\sigma^2}.
\]

\begin{proof}
\begin{eqnarray*}
{\mathbb E}\left[e^{uX}\right]
&=&\int^{\infty}_{-\infty}e^{ux}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\\
&=&\int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2-2\sigma^2ux}{2\sigma^2}}dx\\
&=&\int^{\infty}_{-\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{[x-(\mu+\sigma^2u)]^2-(2\sigma^2u\mu+\sigma^4u^2)}{2\sigma^2}}dx\\
&=&e^{u\mu+\frac{\sigma^2u^2}{2}}\int^{\infty}_{-\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{[x-(\mu+\sigma^2u)]^2}{2\sigma^2}}dx\\
&=&e^{u\mu+\frac{\sigma^2u^2}{2}}
\end{eqnarray*}
\end{proof}

\noindent (ii) Verify that Jensen's inequality holds (as it must):
\[
{\mathbb E}\varphi(X) \ge \varphi({\mathbb E}X).
\]

\begin{proof}${\mathbb E}[\varphi(X)]={\mathbb E}\left[e^{uX}\right]=e^{u\mu+\frac{u^2\sigma^2}{2}}\ge
e^{u\mu}=\varphi({\mathbb E}[X])$.
\end{proof}

\medskip

\noindent  $\blacktriangleright$  {\bf Exercise 1.7.} For each positive integer $n$, define $f_n$ to be the normal density with mean zero and variance $n$, i.e.,
\[
f_n(x) = \frac{1}{\sqrt{2n\pi}} e^{-\frac{x^2}{2n}}.
\]

\smallskip

\noindent (i) What is the function $f(x) = \lim_{n\to\infty}f_n(x)$?

\begin{solution}
Since $|f_n(x)|\le
\frac{1}{\sqrt{2n\pi}}$, $f(x)=\lim_{n\to\infty}f_n(x)=0$.
\end{solution}

\noindent (ii) What is $\lim_{n\to\infty}\int_{-\infty}^{\infty} f_n(x)dx$?

\begin{solution} By the change of variable formula,
$\int_{-\infty}^{\infty}f_n(x)dx=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=1$.
So we must have
\[
\lim_{n\to\infty}\int_{-\infty}^{\infty}f_n(x)dx=1.\]
\end{solution}

\noindent (iii) Note that
\[
\lim_{n\to\infty}\int_{-\infty}^{\infty} f_n(x)dx \ne \int_{-\infty}^{\infty} f(x)dx.
\]
Explain why this does not violate the Monotone Convergence Theorem, Theorem 1.4.5.

\begin{solution} This is not contradictory with the Monotone
Convergence Theorem because $\{f_n\}_{n\ge 1}$ doesn't increase to
$0$.
\end{solution}

\medskip

\noindent  $\blacktriangleright$  {\bf Exercise 1.8. (Moment-generating function).} Let $X$ be a nonnegative random variable, and assume that
\[
\varphi(t) = {\mathbb E}e^{tX}
\]
is finite for every $t \in {\mathbb R}$. Assume further that ${\mathbb E}\left[X e^{tX}\right] < \infty$ for every $t\in {\mathbb R}$. The purpose of this exercise is to show that $\varphi'(t) = {\mathbb E}\left[X e^{tX}\right]$ and, in particular, $\varphi'(0)= {\mathbb E}X$.

We recall the definition of derivative:
\[
\varphi'(t) = \lim_{s\to t}\frac{\varphi(t)-\varphi(s)}{t-s}=\lim_{s\to t}\frac{{\mathbb E} e^{tX} - {\mathbb E}e^{sX}}{t-s} = \lim_{s\to t} {\mathbb E} \left[\frac{e^{tX}-e^{sX}}{t-s}\right].
\]
The limit above is taken over a {\it continuous} variable $s$, but we can choose a sequence of numbers $\{s_n\}_{n=1}^{\infty}$ converging to $t$ and compute
\[
\lim_{s_n\to t} {\mathbb E} \left[\frac{e^{tX}-e^{s_nX}}{t-s_n}\right],
\]
where now we are taking a limit of the expectation of the {\it sequence} of random varibles
\[
Y_n = \frac{e^{tX}-e^{s_nX}}{t-s_n}.
\]
If this limit turns out to be the same, regardless of how we choose the sequence $\{s_n\}_{n=1}^{\infty}$ that converges to $t$, then this limit is also the same as $\lim_{s\to t}{\mathbb E}\left[\frac{e^{tX}-e^{sX}}{t-s}\right]$ and is $\varphi'(t)$.

The Mean Value Theorem from calculus states that if $f(t)$ is a differentiable function, then for any two numbers $s$ and $t$, there is a number $\theta$ between $s$ and $t$ such that
\[
f(t) - f(s) = f'(\theta)(t-s).
\]
If we fix $\omega \in \Omega$  and define $f(t) = e^{tX(\omega)}$, then this becomes
\[
e^{tX(\omega)} - e^{sX(\omega)} = (t-s) X(\omega) e^{\theta(\omega)X(\omega)},\tag{1.9.1}
\]
where $\theta(\omega)$ is a number depending on $\omega$ (i.e., a random variable lying between $t$ and $s$).

\smallskip

\noindent (i) Use the Dominated Convergence Theorem (Theorem 1.4.9) and equation (1.9.1) to show that
\[
\lim_{n\to\infty} {\mathbb E}Y_n = {\mathbb E} \left[\lim_{n\to\infty} Y_n \right] = {\mathbb E} \left[Xe^{tX}\right]. \tag{1.9.2}
\]
This establishes the desired formula $\varphi'(t) = {\mathbb E}\left[Xe^{tX}\right]$.

\begin{proof}
 By (1.9.1),
$|Y_n|=\left|\frac{e^{tX}-e^{s_nX}}{t-s_n}\right|=|Xe^{\theta_n
X}|=Xe^{\theta_n X}\le Xe^{\max\{2|t|,1\} X}$. The last inequality is by $X\ge 0$
and the fact that $\theta_n$ is between $t$ and $s_n$, and hence
smaller than $\max\{2|t|,1\}$ for $n$ sufficiently large. So by the Dominated
Convergence Theorem,
$\varphi'(t)=\lim_{n\to\infty}{\mathbb E}[Y_n]={\mathbb E}[\lim_{n\to\infty}Y_n]={\mathbb E}[Xe^{tX}]$.\end{proof}

\noindent (ii) Suppose the random variable $X$ can take both positive and negative values and ${\mathbb E}e^{tX} < \infty$ and ${\mathbb E}\left[|X|e^{tX}\right] < \infty$ for every $t\in {\mathbb R}$. Show that once again $\varphi'(t) = {\mathbb E}\left[Xe^{tX}\right]$. (Hint: Use the notation (1.3.1) to write $X= X^+ - X^-$.)

\begin{proof}
Since ${\mathbb E}[e^{tX^+}1_{\{X\ge 0\}}]+{\mathbb E}[e^{-tX^-}1_{\{X<
0\}}]={\mathbb E}[e^{tX}]<\infty$ for every $t\in \mathbb R$,
${\mathbb E}[e^{t|X|}]={\mathbb E}[e^{tX^+}1_{\{X\ge 0\}}]+{\mathbb E}[e^{-(-t)X^-}1_{\{X<
0\}}]<\infty$ for every $t\in\mathbb R$. Similarly, we have
${\mathbb E}[|X|e^{t|X|}]<\infty$ for every $t\in\mathbb R$. So, similar to
(i), we have $|Y_n|=|Xe^{\theta_n X}|\le |X|e^{\max\{2t,1\}|X|}$ for $n$
sufficiently large, and by the Dominated Convergence Theorem,
$\varphi'(t)=\lim_{n\to\infty}{\mathbb E}[Y_n]={\mathbb E}[\lim_{n\to\infty}Y_n]={\mathbb E}[Xe^{tX}]$.
\end{proof}


\medskip

\noindent  $\blacktriangleright$  {\bf Exercise 1.9.} Suppose $X$ is a random variable on some probability space $(\Omega, {\cal F}, {\mathbb P})$, $A$ is a set in ${\cal F}$, and for every Borel subset $B$ of ${\mathbb R}$, we have
\[
\int_A 1_B(X(\omega)) d {\mathbb P}(\omega) = {\mathbb P}(A) \cdot {\mathbb P}\{X\in B\}. \tag{1.9.3}
\]
Then we say that $X$ is {\it independent} of the event $A$.

Show that if $X$ is independent of an event $A$, then
\[
\int_A g(X(\omega)) d{\mathbb P}(\omega) = {\mathbb P}(A) \cdot {\mathbb E}g(X)
\]
for every nonnegative, Borel-measurable function $g$.

\begin{proof}
If $g(x)$ is of the form $1_B(x)$, where
$B$ is a Borel subset of $\mathbb R$, then the desired equality is
just (1.9.3). By the linearity of Lebesgue integral, the desired
equality also holds for simple functions, i.e. $g$ of the form
$g(x)=\sum_{i=1}^n1_{B_i}(x)$, where each $B_i$ is a Borel subset of
$\mathbb R$. Since any nonnegative, Borel-measurable function $g$ is
the limit of an increasing sequence of simple functions, the desired
equality can be proved by the Monotone Convergence Theorem.
\end{proof}

\medskip

\noindent  $\blacktriangleright$  {\bf Exercise 1.10.} Let ${\mathbb P}$ be the uniform (Lebesgue) measure on $\Omega = [0,1]$. Define
\[
Z(\omega) =
\begin{cases}
0 & \mbox{if $0\le \omega < \frac{1}{2}$,} \\
2 & \mbox{if $\frac{1}{2}\le \omega \le 1$.} \end{cases}
\]
For $A \in {\cal B}[0,1]$, define
\[
\widetilde {\mathbb P}(A) = \int_A Z(\omega) d {\mathbb P}(\omega).
\]

\smallskip

\noindent (i) Show that $\widetilde {\mathbb P}$ is a probability measure.

\begin{proof}If $\{A_i\}_{i=1}^{\infty}$ is a
sequence of disjoint Borel subsets of $[0,1]$, then by the Monotone
Convergence Theorem, $\widetilde {\mathbb P}(\cup_{i=1}^{\infty}A_i)$ equals
to
\begin{eqnarray*}
\int1_{\cup_{i=1}^{\infty}A_i}Zd{\mathbb P}=\int\lim_{n\to\infty}1_{\cup_{i=1}^nA_i}Zd{\mathbb P}=\lim_{n\to\infty}\int1_{\cup_{i=1}^nA_i}Zd{\mathbb P}=\lim_{n\to\infty}\sum_{i=1}^n\int_{A_i}Zd{\mathbb P}
=\sum_{i=1}^{\infty}\widetilde {\mathbb P}(A_i).
\end{eqnarray*}Meanwhile, $\widetilde
{\mathbb P}(\Omega)=2{\mathbb P}([\frac{1}{2},1])=1$. So $\widetilde {\mathbb P}$ is a probability
measure.
\end{proof}

\noindent (ii) Show that if ${\mathbb P}(A) = 0$, then $\widetilde {\mathbb P}(A) = 0$. We say that $\widetilde {\mathbb P}$ is {\it absolutely continuous} with respect to ${\mathbb P}$.

\begin{proof}If ${\mathbb P}(A)=0$, then $\widetilde
{\mathbb P}(A)=\int_AZd{\mathbb P}=2\int_{A\cap [\frac{1}{2},1]}d{\mathbb P}=2{\mathbb P}(A\cap
[\frac{1}{2},1])=0$.\end{proof}

\noindent (iii) Show that there is a set $A$ for which $\widetilde {\mathbb P}(A) = 0$ but ${\mathbb P}(A) > 0$. In other words, $\widetilde {\mathbb P}$ and ${\mathbb P}$ are not equivalent.

\begin{proof}Let $A=[0,\frac{1}{2})$. \end{proof}


\medskip

\noindent $\blacktriangleright$ {\bf Exercise 1.11.} In Example 1.6.6, we began with a standard normal random variable $X$ under a measure ${\mathbb P}$. According to Exercise 1.6, this random variable has the moment-generating function
\[
{\mathbb E} e^{uX} = e^{\frac{1}{2}u^2} \; \mbox{for all $u\in {\mathbb P}$.}
\]
The moment-generating function of a random variable determines its distribution. In particular, any random variable that has moment-generating function $e^{\frac{1}{2}u^2}$ must be standard normal.

In Example 1.6.6, we also defined $Y=X+\theta$, where $\theta$ is a constant, we set $Z=e^{-\theta X -\frac{1}{2}\theta^2}$, and we defined $\widetilde {\mathbb P}$ by the formula (1.6.9):
\[
\widetilde {\mathbb P}(A) = \int_A Z(\omega) d{\mathbb P}(\omega) \; \mbox{for all $A\in {\cal F}$.}
\]
We showed by considering its cumulative distribution function that $Y$ is a standard normal random variable under $\widetilde {\mathbb P}$. Give another proof that $Y$ is standard normal under $\widetilde {\mathbb P}$ by verifying the moment-generating function formula
\[
\widetilde{\mathbb E}e^{uY} = e^{\frac{1}{2}u^2} \; \mbox{for all $u\in \mathbb R$}.
\]

\begin{proof}
\begin{eqnarray*}
\widetilde {\mathbb E}\left[e^{uY}\right]={\mathbb E}\left[e^{uY}Z\right]={\mathbb E}\left[e^{uX+u\theta}e^{-\theta
X-\frac{\theta^2}{2}}\right]=e^{u\theta-\frac{\theta^2}{2}}E\left[e^{(u-\theta)X}\right]=e^{u\theta-\frac{\theta^2}{2}}e^{\frac{(u-\theta)^2}{2}}=e^{\frac{u^2}{2}}.
\end{eqnarray*}
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 1.12.} In Example 1.6.6, we began with a standard normal random variable $X$ on a probability space $(\Omega, {\cal F}, {\mathbb P})$ and defined the random variable $Y=X+\theta$, where $\theta$ is a constant. We also defined $Z = e^{-\theta X - \frac{1}{2}\theta^2}$ and used $Z$ as the Radon-Nikod\'{y}m derivative to construct the probability measure $\widetilde {\mathbb P}$ by the formula (1.6.9):
\[
\widetilde {\mathbb P}(A) = \int_A Z(\omega) d{\mathbb P}(\omega), \; \mbox{for all $A\in {\cal F}$.}
\]
Under $\widetilde{\mathbb P}$, the random variable $Y$ was shown to be standard normal.

We now have a standard normal random variable $Y$ on the probability space $(\Omega, {\cal F}, \widetilde {\mathbb P})$, and $X$ is related to $Y$ by $X = Y - \theta$. By what we have just stated, with $X$ replaced by $Y$ and $\theta$ replaced by $-\theta$, we could define $\widehat Z = e^{\theta Y - \frac{1}{2}\theta^2}$ and then use $\widehat Z$ as a Radon-Nikod\'{y}m derivative to construct a probability measure $\widehat {\mathbb P}$ by the formula
\[
\widehat {\mathbb P}(A) = \int_A \widehat Z(\omega) d\widetilde {\mathbb P}(\omega) \; \mbox{for all $A\in {\cal F}$,}
\]
so that, under $\widehat {\mathbb P}$, the random variable $X$ is standard normal. Show that $\widehat Z = \frac{1}{Z}$ and $\widehat {\mathbb P} = {\mathbb P}$.

\begin{proof}
First, $\widehat Z=e^{\theta
Y-\frac{\theta^2}{2}}=e^{\theta(X+\theta)-\frac{\theta^2}{2}}=e^{\frac{\theta^2}{2}+\theta
X}=Z^{-1}$. Second, for any $A\in {\cal F}$, $\widehat {\mathbb P}(A)=\int_A\widehat
Zd\widetilde {\mathbb P}=\int(1_A\widehat Z)Zd{\mathbb P}=\int1_Ad{\mathbb P}={\mathbb P}(A)$. So ${\mathbb P}=\widehat {\mathbb P}$. In
particular, $X$ is standard normal under $\widehat {\mathbb P}$, since it's
standard normal under ${\mathbb P}$.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 1.13 (Change of measure for a normal random variable).} A nonrigorous but informative derivation of the formula for the Radon-Nikod\'{y}m derivative $Z(\omega)$ in Example 1.6.6 is provided by this exercise. As in that example, let $X$ be a standard normal random variable on some probability space $(\Omega, {\cal F}, {\mathbb P})$, and let $Y=X+\theta$. Our goal is to define a strictly positive random variable $Z(\omega)$ so that when we set
\[
\widetilde {\mathbb P}(A) = \int_A Z(\omega) d\widetilde {\mathbb P}(\omega) \; \mbox{for all $A\in {\cal F}$}, \tag{1.9.4}
\]
the random variable $Y$ under $\widetilde {\mathbb P}$ is standard normal. If we fix $\overline{\omega}\in \Omega$ and choose a set $A$ that contains $\overline{\omega}$ and is ``small," then (1.9.4) gives
\[
\widetilde {\mathbb P}(A) \approx Z(\overline{\omega}){\mathbb P}(A),
\]
where the symbol $\approx$ means ``is approximately equal to." Dividing by ${\mathbb P}(A)$, we see that
\[
\frac{\widetilde {\mathbb P}(A)}{{\mathbb P}(A)} \approx Z(\overline{\omega})
\]
for ``small" sets $A$ containing $\overline{\omega}$. We use this observation to identify $Z(\overline{\omega})$.

With $\overline{\omega}$ fixed, let $x=X(\overline{\omega})$. For $\epsilon>0$, we define $B(x,\epsilon)=\left[x-\frac{\epsilon}{2}, x+\frac{\epsilon}{2}\right]$ to be the closed interval centered at $x$ and having length $\epsilon$. Let $y=x+\theta$ and $B(y,\epsilon)=\left[y-\frac{\epsilon}{2}, y+\frac{\epsilon}{2}\right]$.

\smallskip

\noindent (i) Show that
\[
\frac{1}{\epsilon} {\mathbb P}\{X \in B(x,\epsilon) \} \approx \frac{1}{\sqrt{2\pi}} \exp \left\{-\frac{X^2(\overline{\omega})}{2}\right\}.
\]

\begin{proof}
\[
\frac{1}{\epsilon}{\mathbb P}(X\in
B(x,\epsilon))=\frac{1}{\epsilon}\int_{x-\frac{\epsilon}{2}}^{x+\frac{\epsilon}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{u^2}{2}}du
\approx
\frac{1}{\epsilon}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\cdot
\epsilon=\frac{1}{\sqrt{2\pi}}e^{-\frac{X^2(\bar\omega)}{2}}.
\]
\end{proof}

\noindent (ii) In order for $Y$ to be a standard normal random variable under $\widetilde{\mathbb P}$, show that we must have
\[
\frac{1}{\epsilon} \widetilde {\mathbb P}\{Y \in B(y, \epsilon) \} \approx \frac{1}{\sqrt{2\pi}} \exp \left\{-\frac{Y^2(\overline{\omega})}{2}\right\}.
\]

\begin{proof} Similar to (i).\end{proof}

\noindent (iii) Show that $\{X\in B(x,\epsilon)\}$ and $\{Y\in B(y,\epsilon)\}$ are the same set, which we call $A(\overline{\omega},\epsilon)$. This set contains $\overline{\omega}$ and is ``small" when $\epsilon>0$ is small.

\begin{proof} $\{X\in B(x,\epsilon)\}=\{X\in
B(y-\theta,\epsilon)\}=\{X+\theta\in B(y,\epsilon)\}=\{Y\in
B(y,\epsilon)\}$.
\end{proof}

\noindent (iv) Show that
\[
\frac{\widetilde {\mathbb P}(A)}{{\mathbb P}(A)} \approx \exp \left\{-\theta X(\overline{\omega})-\frac{1}{2}\theta^2\right\}.
\]
The right-hand side is the value we obtained for $Z(\overline{\omega})$ in Example 1.6.6.

\begin{proof} By (i)-(iii), $\frac{\widetilde {\mathbb P}(A)}{{\mathbb P}(A)}$ is
approximately
\[
\frac{\frac{\epsilon}{\sqrt{2\pi}}e^{-\frac{Y^2(\bar\omega)}{2}}}{\frac{\epsilon}{\sqrt{2\pi}}e^{-\frac{X^2(\bar\omega)}{2}}}=e^{-\frac{Y^2(\bar\omega)-X^2(\bar\omega)}{2}}
=e^{-\frac{(X(\bar\omega)+\theta)^2-X^2(\bar\omega)}{2}}=e^{-\theta
X(\bar\omega)-\frac{\theta^2}{2}}.
\]
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 1.14 (Change of measure for an exponential random variable).} Let $X$ be a nonnegative random variable defined on a probability space $(\Omega, {\cal F}, {\mathbb P})$ with the {\it exponential distribution}, which is
\[
{\mathbb P}\{X\le a \} = 1 - e^{-\lambda a}, \; a\ge 0,
\]
where $\lambda$ is a positive constant. Let $\tilde \lambda$ be another positive constant, and define
\[
Z = \frac{\tilde\lambda}{\lambda} e^{-(\tilde \lambda - \lambda)X}
\]
Define $\widetilde {\mathbb P}$ by
\[
\widetilde {\mathbb P}(A) = \int_A Z d{\mathbb P} \; \mbox{for all $A\in {\cal F}$.}
\]

\smallskip

\noindent (i) Show that $\widetilde {\mathbb P}(\Omega)=1$.

\begin{proof}
\[
\widetilde
{\mathbb P}(\Omega)=\int\frac{\widetilde\lambda}{\lambda}e^{-(\widetilde
\lambda-\lambda)X}d{\mathbb P}=\frac{\widetilde
\lambda}{\lambda}\int_0^{\infty}e^{-(\widetilde\lambda-\lambda)x}\lambda
e^{-\lambda x}dx=\int_0^{\infty}\widetilde\lambda
e^{-\widetilde\lambda x}dx=1.
\]
\end{proof}

\noindent (ii) Compute the cumulative distribution function
\[
\widetilde {\mathbb P}\{X\le a\} \; \mbox{for $a\ge 0$}
\]
for the random variable $X$ under the probability measure $\widetilde {\mathbb P}$.

\begin{solution}
\[
\widetilde {\mathbb P}(X\le a)=\int_{\{X\le a\}}\frac{\widetilde
\lambda}{\lambda}e^{-(\widetilde\lambda-\lambda)X}d{\mathbb P}=\int_0^a\frac{\widetilde\lambda}{\lambda}e^{-(\widetilde
\lambda-\lambda)x}\lambda e^{-\lambda x}dx=\int_0^a\widetilde\lambda
e^{-\widetilde\lambda x}dx=1-e^{-\widetilde\lambda a}.\]
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 1.15 (Provided by Alexander Ng).} Let $X$ be a random variable on a probability space $(\Omega, {\cal F}, {\mathbb P})$, and assume $X$ has a density function $f(x)$ that is positive for every $x\in {\mathbb R}$. Let $g$ be a strictly increasing, differentiable function satisfying
\[
\lim_{y\to -\infty} g(y) = -\infty, \lim_{y\to\infty}g(y) = \infty,
\]
and define the random variable $Y=g(X)$.

Let $h(y)$ be an arbitrary nonnegative function satisfying $\int_{-\infty}^{\infty} h(y)dy = 1$. We want to change the probability measure so that $h(y)$ is the density function for the random variable $Y$. To do this, we define
\[
Z = \frac{h(g(X))g'(X)}{f(X)}.
\]

\smallskip

\noindent (i) Show that $Z$ is nonnegative and ${\mathbb E}Z=1$. Now define $\widetilde {\mathbb P}$ by
\[
\widetilde {\mathbb P}(A) = \int_A Z d{\mathbb P} \; \mbox{for all $A\in {\cal F}$.}
\]

\begin{proof}Clearly $Z\ge 0$. Furthermore, we
have
\[
{\mathbb E} \{Z\}={\mathbb E}\left\{\frac{h(g(X))g'(X)}{f(X)}\right\}=\int_{-\infty}^{\infty}\frac{h(g(x))g'(x)}{f(x)}f(x)dx=\int_{-\infty}^{\infty}h(g(x))dg(x)=\int_{-\infty}^{\infty}h(u)du=1.
\]
\end{proof}

\noindent (ii) Show that $Y$ has density $h$ under $\widetilde {\mathbb P}$.

\begin{proof}
\[
\widetilde {\mathbb P}(Y\le a)=\int_{\{g(X)\le
a\}}\frac{h(g(X))g'(X)}{f(X)}d{\mathbb P}=\int_{-\infty}^{g^{-1}(a)}\frac{h(g(x))g'(x)}{f(x)}f(x)dx=\int_{-\infty}^{g^{-1}(a)}h(g(x))dg(x).\]
By the change of variable formula, the last equation above equals to
$\int^a_{-\infty}h(u)du$. So $Y$ has density $h$ under $\widetilde
{\mathbb P}$.
\end{proof}



\section{Information and Conditioning}


\noindent $\blacktriangleright$  {\bf Exercise 2.1.} Let $(\Omega, {\cal F}, {\mathbb P})$ be a general probability space, and suppose a random variable $X$ on this space is measurable with respect to the trivial $\sigma$-algebra ${\cal F}_0 = \{\emptyset, \Omega\}$. Show that $X$ is not random (i.e., there is a constant $c$ such that $X(\omega)=c$ for all $\omega\in \Omega$). Such a random variable is called {\it degenerate}.

\begin{proof}For any real number $a$, we have $\{X\le
a\}\in {\cal F}_0=\{\emptyset,\Omega\}$. So ${\mathbb P}(X\le a)$ is either
$0$ or $1$. Since $\lim_{a\to\infty}{\mathbb P}(X\le a)=1$ and
$\lim_{a\to -\infty}{\mathbb P}(X\le a)=0$, we can find a number $x_0$ such
that ${\mathbb P}(X\le x_0)=1$ and ${\mathbb P}(X\le x)=0$ for any $x<x_0$. So
\[
{\mathbb P}(X=x_0)=\lim_{n\to\infty}{\mathbb P}\left(x_0-\frac{1}{n}<X\le
x_0\right)=\lim_{n\to\infty}\left({\mathbb P}(X\le x_0)-{\mathbb P}\left(X\le x_0-\frac{1}{n}\right)\right)=1.\]
Since $\{X=x_0\} \in \{\emptyset, \Omega\}$, we conclude $\{X=x_0\}=\Omega$.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 2.2.} Independence of random variables can be affected by changes of measure. To illustrate this point, consider the space of two coin tosses $\Omega_2 = \{ HH, HT, TH, TT\}$, and let stock prices be given by
\begin{eqnarray*}
& & S_0 = 4, S_1(H) = 8, S_1(T) = 2, \\
& & S_2(HH) = 16, S_2(HT) = S_2(TH) = 4, S_2(TT) = 1.
\end{eqnarray*}
Consider the two probability measures given by
\begin{eqnarray*}
&& \widetilde {\mathbb P}(HH) = \frac{1}{4}, \widetilde {\mathbb P}(HT) = \frac{1}{4}, \widetilde {\mathbb P}(TH)=\frac{1}{4}, \widetilde {\mathbb P}(TT) = \frac{1}{4}, \\
&& {\mathbb P}(HH) = \frac{4}{9}, {\mathbb P}(HT) = \frac{2}{9}, {\mathbb P}(TH) = \frac{2}{9}, {\mathbb P}(TT) = \frac{1}{9}.
\end{eqnarray*}
Define the random variable
\[
X = \begin{cases}
1 & \mbox{if $S_2 = 4$,} \\
0 & \mbox{if $S_2 \ne 4$.}
\end{cases}
\]

\smallskip

\noindent (i) List all the sets in $\sigma(X)$.

\begin{solution}$\sigma(X)=\{\emptyset, \Omega,
\{HT, TH\},\{TT, HH\}\}$. \end{solution}

\noindent (ii) List all the sets in $\sigma(S_1)$.

\begin{solution}
$\sigma(S_1)=\{\emptyset,\Omega,\{HH,HT\},\{TH,TT\}\}$.
\end{solution}

\noindent (iii) Show that $\sigma(X)$ and $\sigma(S_1)$ are independent under the probability measure $\widetilde {\mathbb P}$.

\begin{proof} $\widetilde {\mathbb P}(\{HT,TH\}\cap\{HH,HT\})=\widetilde
{\mathbb P}(\{HT\})=\frac{1}{4}$, $\widetilde {\mathbb P}(\{HT,TH\})=\widetilde
{\mathbb P}(\{HT\})+\widetilde {\mathbb P}(\{TH\})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$,
and $\widetilde {\mathbb P}(\{HH,HT\})=\widetilde {\mathbb P}(\{HH\})+\widetilde
{\mathbb P}(\{HT\})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$. So we have
\[
\widetilde {\mathbb P}(\{HT, TH\}\cap \{HH,HT\})=\widetilde {\mathbb P}(\{HT,
TH\})\widetilde {\mathbb P}(\{HH,HT\}).
\]
Similarly, we can work on other elements of $\sigma(X)$ and
$\sigma(S_1)$ and show that $\widetilde {\mathbb P}(A\cap B)=\widetilde
{\mathbb P}(A)\widetilde {\mathbb P}(B)$ for any $A\in \sigma(X)$ and $B\in
\sigma(S_1)$. So $\sigma(X)$ and $\sigma(S_1)$ are independent under
$\widetilde {\mathbb P}$.
\end{proof}

\noindent (iv) Show that $\sigma(X)$ and $\sigma(S_1)$ are not independent under the probability measure ${\mathbb P}$.

\begin{proof} ${\mathbb P}(\{HT,TH\}\cap
\{HH,HT\})={\mathbb P}(\{HT\})=\frac{2}{9}$,
${\mathbb P}(\{HT,TH\})=\frac{2}{9}+\frac{2}{9}=\frac{4}{9}$ and
${\mathbb P}(\{HH,HT\})=\frac{4}{9}+\frac{2}{9}=\frac{6}{9}$. So
\[
{\mathbb P}(\{HT,TH\}\cap \{HH,HT\})\ne {\mathbb P}(\{HT,TH\}){\mathbb P}(\{HH,HT\}).
\]
Hence $\sigma(X)$ and $\sigma(S_1)$ are not independent under ${\mathbb P}$.
\end{proof}

\noindent (v) Under ${\mathbb P}$, we have ${\mathbb P}\{S_1=8\}=\frac{2}{3}$ and ${\mathbb P}\{S_1=2\}=\frac{1}{3}$. Explain intuitively why, if you are told that $X=1$, you would want to revise your estimate of the distribution of $S_1$.

\begin{solution} Because $S_1$ and $X$ are not independent under
the probability measure ${\mathbb P}$, knowing the value of $X$ will affect
our opinion on the distribution of $S_1$.\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 2.3 (Rotating the axes).} Let $X$ and $Y$ be independent standard normal random variables. Let $\theta$ be a constant, and define random variables
\[
V = X\cos\theta+Y\sin\theta \;\mbox{and}\; W = -X\sin\theta+Y\cos\theta.
\]
Show that $V$ and $W$ are independent standard normal random variables.

\begin{proof}We note $(V,W)$ are jointly Gaussian. In order to prove their independence, it suffices to show they are
uncorrelated. Indeed,
\begin{eqnarray*}
{\mathbb E}[VW] &=& {\mathbb E}[-X^2\sin\theta\cos\theta+XY\cos^2\theta-XY\sin^2\theta+Y^2\sin\theta\cos\theta] \\
&=& -\sin\theta\cos\theta+0+0+\sin\theta\cos\theta \\
&=&0.
\end{eqnarray*}
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 2.4.} In Example 2.2.8, $X$ is a standard normal random variable and $Z$ is an independent random variable satisfying
\[
{\mathbb P}\{Z=1\} = {\mathbb P}\{Z=-1\} = \frac{1}{2}.
\]
We defined $Y=XZ$ and showed that $Y$ is standard normal. We established that although $X$ and $Y$ are uncorrelated, they are not independent. In this exercise, we use moment-generating functions to show that $Y$ is standard normal and $X$ and $Y$ are not independent.

\smallskip

\noindent (i) Establish the joint moment-generating function formula
\[
{\mathbb E} e^{uX+vY} = e^{\frac{1}{2}(u^2+v^2)} \cdot \frac{e^{uv}+e^{-uv}}{2}.
\]

\begin{solution}
\begin{eqnarray*}
{\mathbb E}\left[e^{uX+vY}\right]
&=&{\mathbb E}\left[e^{uX+vXZ}\right]\\
&=&{\mathbb E}\left[e^{uX+vXZ}|Z=1\right]P(Z=1)+{\mathbb E}\left[e^{uX+vXZ}|Z=-1\right]P(Z=-1)\\
&=&\frac{1}{2}{\mathbb E}\left[e^{uX+vX}\right]+\frac{1}{2}E\left[e^{uX-vX}\right]\\
&=&\frac{1}{2} \left[ e^{\frac{(u+v)^2}{2}}+e^{\frac{(u-v)^2}{2}} \right]\\
&=&e^{\frac{u^2+v^2}{2}} \cdot \frac{e^{uv}+e^{-uv}}{2}.
\end{eqnarray*}
\end{solution}

\noindent (ii) Use the formula above to show that ${\mathbb E}e^{vY} = e^{\frac{1}{2}v^2}$. This is the moment-generating function for a standard normal random variable, and thus $Y$ must be a standard normal random variable.

\begin{proof} Let $u=0$, we are done.
\end{proof}

\noindent (iii) Use the formula in (i) and Theorem 2.2.7(iv) to show that $X$ and $Y$ are not independent.

\begin{proof}${\mathbb E}\left[e^{uX}\right]=e^{\frac{u^2}{2}}$ and
${\mathbb E}\left[e^{vY}\right]=e^{\frac{v^2}{2}}$. So ${\mathbb E}\left[e^{uX+vY}\right] \ne
{\mathbb E}\left[e^{uX}\right]{\mathbb E}\left[e^{vY}\right]$. Therefore $X$ and $Y$ cannot be
independent.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 2.5.} Let $(X, Y)$ be a pair of random variables with joint density function
\[
f_{X,Y}(x,y)=\begin{cases}
\frac{2|x|+y}{\sqrt{2\pi}}\exp\left\{-\frac{(2|x|+y)^2}{2}\right\} & \mbox{if $y\ge -|x|$,} \\
0 & \mbox{if $y<-|x|$.}
\end{cases}
\]
Show that $X$ and $Y$ are standard normal random variables and that they are uncorrelated but not independent.

 \begin{proof} The density $f_X(x)$ of $X$ can be
obtained by
\[
f_X(x)=\int f_{X,Y}(x,y)dy=\int_{\{y\ge
-|x|\}}\frac{2|x|+y}{\sqrt{2\pi}}e^{-\frac{(2|x|+y)^2}{2}}dy=\int_{\{\xi\ge|x|\}}\frac{\xi}{\sqrt{2\pi}}e^{-\frac{\xi^2}{2}}d\xi=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}.
\]
The density $f_Y(y)$ of $Y$ can be obtained by \begin{eqnarray*}
f_Y(y)&=&\int f_{XY}(x,y)dx\\
&=&\int 1_{\{|x|\ge
-y\}}\frac{2|x|+y}{\sqrt{2\pi}}e^{-\frac{(2|x|+y)^2}{2}}dx\\
&=&\int^{\infty}_{0\vee(-y)}\frac{2x+y}{\sqrt{2\pi}}e^{-\frac{(2x+y)^2}{2}}dx+\int_{-\infty}^{0\wedge
y}\frac{-2x+y}{\sqrt{2\pi}}e^{-\frac{(-2x+y)^2}{2}}dx\\
&=&\int^{\infty}_{0\vee(-y)}\frac{2x+y}{\sqrt{2\pi}}e^{-\frac{(2x+y)^2}{2}}dx+\int_{\infty}^{0\vee(-y)}\frac{2x+y}{\sqrt{2\pi}}e^{-\frac{(2x+y)^2}{2}}d(-x)\\
&=&2\int_{|y|}^{\infty}\frac{\xi}{\sqrt{2\pi}}e^{-\frac{\xi^2}{2}}d(\frac{\xi}{2})\\
&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}.
\end{eqnarray*}
So both $X$ and $Y$ are standard normal random variables. Since
$f_{X,Y}(x,y)\ne f_X(x)f_Y(y)$, $X$ and $Y$ are not independent.
However, if we set
$F(t)=\int_t^{\infty}\frac{u^2}{\sqrt{2\pi}}e^{-\frac{u^2}{2}}du$,
we have
\begin{eqnarray*}
{\mathbb E}\left[XY\right]
&=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} xyf_{X,Y}(x,y)dxdy\\
&=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} xy1_{\{y\ge
-|x|\}}\frac{2|x|+y}{\sqrt{2\pi}}e^{-\frac{(2|x|+y)^2}{2}}dxdy\\
&=&\int_{-\infty}^{\infty}
xdx\int^{\infty}_{-|x|}y\frac{2|x|+y}{\sqrt{2\pi}}e^{-\frac{(2|x|+y)^2}{2}}dy\\
&=&\int_{-\infty}^{\infty}
xdx\int^{\infty}_{|x|}(\xi-2|x|)\frac{\xi}{\sqrt{2\pi}}e^{-\frac{\xi^2}{2}}d\xi\\
&=&\int_{-\infty}^{\infty}xdx \left(\int^{\infty}_{|x|}\frac{\xi^2}{\sqrt{2\pi}}e^{-\frac{\xi^2}{2}}d\xi-2|x|\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}} \right)\\
&=&\int_0^{\infty}x\int_x^{\infty}\frac{\xi^2}{\sqrt{2\pi}}e^{-\frac{\xi^2}{2}}d\xi
dx+\int^0_{-\infty}x\int^{\infty}_{-x}\frac{\xi^2}{\sqrt{2\pi}}e^{-\frac{\xi^2}{2}}d\xi
dx\\
&=&\int_0^{\infty}xF(x)dx+\int^0_{-\infty}xF(-x)dx \\
&=& 0.
\end{eqnarray*}
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 2.6.} Consider a probability space $\Omega$ with four elements, which we call $a$, $b$, $c$, and $d$ (i.e., $\Omega = \{a, b, c, d\}$). The $\sigma$-algebra ${\cal F}$ is the collection of all subsets of $\Omega$; i.e., the sets in ${\cal F}$ are
\begin{eqnarray*}
& & \Omega, \{a, b, c\}, \{a, b, d\}, \{a, c, d\}, \{b, c, d\}, \\
& & \{a, b\}, \{a, c\}, \{a, d\}, \{b, c\}, \{b, d\}, \{c, d\}, \\
& & \{a\}, \{b\}, \{c\}, \{d\}, \emptyset.
\end{eqnarray*}
We define a probability measure ${\mathbb P}$ by specifying that
\[
{\mathbb P}\{a\} = \frac{1}{6}, {\mathbb P}\{b\}=\frac{1}{3}, {\mathbb P}\{c\} = \frac{1}{4}, {\mathbb P}\{d\}=\frac{1}{4},
\]
and, as usual, the probability of every other set in ${\cal F}$ is the sum of the probabilities of the elements in the set, e.g., ${\mathbb P}\{a, b, c\} = {\mathbb P}\{a\} + {\mathbb P}\{b\} + {\mathbb P}\{c\} = \frac{3}{4}$.

We next define two random variables, $X$ and $Y$, by the formulas
\begin{eqnarray*}
&& X(a) = 1, X(b) = 1, X(c) = -1, X(d) = -1, \\
&& Y(a) = 1, Y(b) = -1, Y(c) = 1, Y(d) = -1.
\end{eqnarray*}
We then define $Z = X + Y$.

\smallskip

\noindent (i) List the sets in $\sigma(X)$.

\begin{solution}
$\sigma(X)=\{\emptyset,\Omega,\{a,b\},\{c,d\}\}$.\end{solution}

\noindent (ii) Determine ${\mathbb E}[Y|X]$ (i.e., specify the values of this random variable for $a$, $b$, $c$, and $d$). Verify that the partial-averaging property is satisfied.

\begin{solution}
\begin{eqnarray*}
{\mathbb E}[Y|X]
&=& \sum_{\alpha\in\{1,-1\}} {\mathbb E}[Y|X=\alpha] 1_{\{X=\alpha\}} \\
&=& \sum_{\alpha\in\{1,-1\}}\frac{{\mathbb E}[Y1_{\{X=\alpha\}}]}{{\mathbb P}(X=\alpha)}1_{\{X=\alpha\}} \\
&=& \frac{1\cdot {\mathbb P}(Y=1,X=1)-1 \cdot {\mathbb P}(Y=-1,X=1)}{{\mathbb P}(X=1)} 1_{\{X=1\}} \\
& & + \frac{1\cdot {\mathbb P}(Y=1,X=-1)-1 \cdot {\mathbb P}(Y=-1,X=-1)}{{\mathbb P}(X=-1)} 1_{\{X=-1\}} \\
&=& \frac{1\cdot {\mathbb P}(\{a\})-1 \cdot {\mathbb P}(\{b\})}{{\mathbb P}(\{a,b\})} 1_{\{X=1\}} + \frac{1\cdot {\mathbb P}(\{c\})-1 \cdot {\mathbb P}(\{d\})}{{\mathbb P}(\{c,d\})} 1_{\{X=-1\}} \\
&=& -\frac{1}{3} 1_{\{X=1\}}.
\end{eqnarray*}
To verify the partial-averaging property, we note
\[
{\mathbb E}\left[{\mathbb E}[Y|X]1_{\{X=1\}}\right] = -\frac{1}{3} {\mathbb E}\left[1_{\{X=1\}}\right] =  -\frac{1}{3} {\mathbb P}(\{a, b\}) = -\frac{1}{6}
\]
and
\[
{\mathbb E}\left[Y1_{\{X=1\}}\right] = {\mathbb P}(X=Y=1) - {\mathbb P}(Y=-1,X=1) = {\mathbb P}(\{a\}) - {\mathbb P}(\{b\}) = -\frac{1}{6}.
\]
Similarly,
\[
{\mathbb E}\left[{\mathbb E}[Y|X]1_{\{X=-1\}}\right] = 0
\]
and
\[
{\mathbb E}\left[Y1_{\{X=-1\}}\right] = {\mathbb P}(Y=1, X=-1) - {\mathbb P}(Y=-1,X=-1) = {\mathbb P}(\{c\}) - {\mathbb P}(\{d\}) = 0.
\]
Together, we can conclude the partial-averaging property hodls.
\end{solution}

\noindent (iii) Determine ${\mathbb E}[Z|X]$. Again, verify the partial-averaging property.

\begin{solution}
\[
{\mathbb E}[Z|X]=X+{\mathbb E}[Y|X]=\frac{2}{3}1_{\{X=1\}} - 1_{\{X=-1\}}.
\]
Verification of the partial-averaging property is skipped.
\end{solution}

\noindent (iv) Compute ${\mathbb E}[Z|X] - {\mathbb E}[Y|X]$. Citing the appropriate properties of conditional expectation from Theorem 2.3.2, explain why you get $X$.

\begin{solution}${\mathbb E}[Z|X]-{\mathbb E}[Y|X]={\mathbb E}[Z-Y|X]={\mathbb E}[X|X]=X$, where the first equality is due to linearity (Theorem 2.3.2(i)) and the last equality is due to ``taking out what is known" (Theorem 2.3.2(ii)). \end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 2.7.} Let $Y$ be an integrable random variable on a probability space $(\Omega, {\cal F}, {\mathbb P})$ and let ${\cal G}$ be a sub-$\sigma$-algebra of ${\cal F}$. Based on the information in ${\cal G}$, we can form the estimate ${\mathbb E}[Y|{\cal G}]$ of $Y$ and define the error of the estimation $\mbox{Err} = Y - {\mathbb E}[Y|{\cal G}]$. This is a random variable with expectation zero and some variance Var(Err). Let $X$ be some other ${\cal G}$-measurable random variable, which we can regard as another estimate of $Y$. Show that
\[
\mbox{Var(Err)} \le \mbox{Var}(Y-X).
\]
In other words, the estimate ${\mathbb E}[Y|{\cal G}]$ minimizes the variance of the error among all estimates based on the information in ${\cal G}$. (Hint: Let $\mu = {\mathbb E}(Y-X)$. Compute the variance of $Y-X$ as
\[
{\mathbb E}[(Y-X-\mu)^2] = {\mathbb E} \left[\left((Y-{\mathbb E}[Y|{\cal G}])+({\mathbb E}[Y|{\cal G}]-X-\mu)\right)^2\right].
\]
Multiply out the right-hand side and use iterated conditioning to show the cross-term is zero.)

 \begin{proof}Let $\mu={\mathbb E}[Y-X]$ and
$\xi={\mathbb E}[Y-X-\mu|{\cal G}]$. Note $\xi$ is ${\cal G}$-measurable, we
have
\begin{eqnarray*}
\mbox{Var}(Y-X)
&=&{\mathbb E}[(Y-X-\mu)^2]\\
&=&{\mathbb E}\left\{\left[(Y-E[Y|{\cal G}])+({\mathbb E}[Y|{\cal G}]-X-\mu)\right]^2\right\}\\
&=&\mbox{Var}(\mbox{Err})+2{\mathbb E}\left[(Y-{\mathbb E}[Y|{\cal G}])\xi\right]+E[\xi^2]\\
&=&\mbox{Var}(\mbox{Err})+2{\mathbb E}\left[Y\xi-{\mathbb E}[Y\xi|{\cal G}]\right]+E[\xi^2]\\
&=&\mbox{Var}(\mbox{Err})+{\mathbb E}[\xi^2]\\
&\ge&\mbox{Var}(\mbox{Err}).
\end{eqnarray*}

\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 2.8.} Let $X$ and $Y$ be integrable random variables on a probability space $(\Omega, {\cal F}, {\mathbb P})$. Then $Y=Y_1+Y_2$, where $Y_1={\mathbb E}[Y|X]$ is $\sigma(X)$-measurable and $Y_2=Y-{\mathbb E}[Y|X]$. Show that $Y_2$ and $X$ are uncorrelated. More generally, show that $Y_2$ is uncorrelated with every $\sigma(X)$-measurable random variable.

\begin{proof} It suffices to prove the more general
case. For any $\sigma(X)$-measurable random variable $\xi$,
${\mathbb E}[Y_2\xi]={\mathbb E}[(Y-E\{Y|X\})\xi]={\mathbb E}[Y\xi-E\{Y|X\}\xi]={\mathbb E}[Y\xi]-{\mathbb E}[Y\xi]=0$.

\end{proof}

\medskip

\noindent $\blacktriangle$ {\bf Exercise 2.9.} Let $X$ be a random variable.

\smallskip

\noindent (i) Give an example of a probability space $(\Omega, {\cal F}, {\mathbb P})$, a random variable $X$ defined on this probability space, and a function $f$ so that the $\sigma$-algebra generated by $f(X)$ is not the trivial $\sigma$-algebra $\{\emptyset,\Omega\}$ but is strictly smaller than the $\sigma$-algebra generated by $X$.

\begin{solution}Consider the dice-toss space similar
to the coin-toss space. Then a typical element $\omega$ in this
space is an infinite sequence $\omega_1\omega_2\omega_3\cdots$, with
$\omega_i\in \{1,2,\cdots, 6\}$ $(i\in\mathbb N)$. We define
$X(\omega)=\omega_1$ and $f(x)=1_{\{\mbox{odd integers}\}}(x)$. Then
it's easy to see
\begin{eqnarray*}
\sigma(X)=\{\emptyset, \Omega, \{\omega:\omega_1=1\}, \cdots,
\{\omega:\omega_1=6\}\}  \end{eqnarray*} and $\sigma(f(X))$ equals
to
\begin{eqnarray*}
\{\emptyset,\Omega,\{\omega:\omega_1=1\}\cup
\{\omega:\omega_1=3\}\cup \{\omega:\omega_1=5\},
\{\omega:\omega_1=2\}\cup \{\omega:\omega_1=4\}\cup
\{\omega:\omega_1=6\}\}.
\end{eqnarray*}
So $\{\emptyset,\Omega\} \subsetneq \sigma(f(X))  \subsetneq
\sigma(X)$, and each of these containment is strict.
\end{solution}

\noindent (ii) Can the $\sigma$-algebra generated by $f(X)$ ever be strictly larger than the $\sigma$-algebra generated by $X$?

\begin{solution}No. $\sigma(f(X))\subset \sigma(X)$ is always
true.\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 2.10.} Let $X$ and $Y$ be random variable (on some unspecified probability space $(\Omega, {\cal F}, {\mathbb P})$), assume they have a joint density $f_{X,Y}(x,y)$, and assume ${\mathbb E}|Y| < \infty$. In particular, for every Borel subset $C$ of ${\mathbb R}^2$, we have
\[
{\mathbb P}\{(X,Y)\in C\} = \int_C f_{X,Y}(x,y)dxdy.
\]

In elementary probability, one learns to compute ${\mathbb E}[Y|X=x]$, which is a {\it nonrandom} function of the {\it dummy variable} $x$, by the formula
\[
{\mathbb E}[Y|X=x] = \int_{-\infty}^{\infty} yf_{Y|X}(y|x)dy, \tag{2.6.1}
\]
where $f_{Y|X}(y|x)$ is the {\it conditional density} defined by
\[
f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_X(x)}.
\]
The denominator in this expression, $f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,\eta)d\eta$, is the {\it marginal density} of $X$, and we must assume it is strictly positive for every $x$. We introduce the symbol $g(x)$ for the function ${\mathbb E}[Y|X=x]$ defined by (2.6.1); i.e.,
\[
g(x) = \int_{-\infty}^{\infty} y f_{Y|X}(y|x)dy = \int_{-\infty}^{\infty} \frac{y f_{X,Y}(x,y)}{f_X(x)}dy
\]

In measure-theoretic probability, conditional expectation is a {\it random variable} ${\mathbb E}[Y|X]$. This exercise is to show that when there is a joint density for $(X,Y)$, this random variable can be obtained by substituting the random variable $X$ in place of the dummy variable $x$ in the function $g(x)$. In other words, this exercise is to show that
\[
{\mathbb E}[Y|X] = g(X).
\]
(We introduced the symbol $g(x)$ in order to avoid the mathematically confusing expression ${\mathbb E}[Y|X=X]$.)

Since $g(X)$ is obviously $\sigma(X)$-measurable, to verify that ${\mathbb E}[Y|X] = g(X)$, we need only check that the partial-averaging property is satisfied. For every Borel-measurable function $h$ mapping ${\mathbb R}$ to ${\mathbb R}$ and satisfying ${\mathbb E}|h(X)| < \infty$, we have
\[
{\mathbb E}h(X) = \int_{-\infty}^{\infty} h(x) f_X(x)dx. \tag{2.6.2}
\]
This is Theorem 1.5.2 in Chapter 1. Similarly, if $h$ is a function of both $x$ and $y$, then
\[
{\mathbb E}h(X,Y) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} h(x,y) f_{X,Y}(x,y)dxdy \tag{2.6.3}
\]
whenever $(X,Y)$ has a joint density $f_{X,Y}(x,y)$. You may use both (2.6.2) and (2.6.3) in your solution to this problem.

Let $A$ be a set in $\sigma(X)$. By the definition of $\sigma(X)$, there is a Borel subset $B$ of ${\mathbb R}$ such that $A=\{\omega\in\Omega; X(\omega)\in B\}$ or, more simply, $A=\{X\in B\}$. Show the partial-averaging property
\[
\int_A g(X) d{\mathbb P} = \int_A Yd{\mathbb P}.
\]
\begin{proof}
\begin{eqnarray*}
\int_Ag(X)dP&=&{\mathbb E}[g(X)1_B(X)] = \int_{-\infty}^{\infty}g(x)1_B(x)f_X(x)dx = \int\int\frac{yf_{X,Y}(x,y)}{f_X(x)}dy1_B(x)f_X(x)dx\\
&=&\int\int
y1_B(x)f_{X,Y}(x,y)dxdy
={\mathbb E}[Y1_B(X)]
={\mathbb E}[YI_A]
=\int_AYd{\mathbb P}.
\end{eqnarray*}

\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 2.11.}

\smallskip

\noindent (i) Let $X$ be a random variable on a probability space $(\Omega, {\cal F}, {\mathbb P})$, and let $W$ be a nonnegative $\sigma(X)$-measurable random variable. Show there exists a function $g$ such that $W=g(X)$. (Hint: Recall that every set in $\sigma(X)$ is of the form $\{X\in B\}$ for some Borel set $B\subset {\mathbb R}$. Suppose first that $W$ is the indicator of such a set, and then use the standard machine.)

\begin{proof} We can find a sequence
$\{W_n\}_{n\ge 1}$ of $\sigma(X)$-measurable simple functions such
that $W_n\uparrow W$. Each $W_n$ can be written as the form
$\sum_{i=1}^{K_n}a_i^n1_{A_i^n}$, where $A_i^n$'s belong to
$\sigma(X)$ and are disjoint. So each $A_i^n$ can be written as
$\{X\in B_i^n\}$ for some Borel subset $B_i^n$ of $\mathbb R$, i.e.
$W_n=\sum_{i=1}^{K_n}a_i^n1_{\{X\in
B_i^n\}}=\sum_{i=1}^{K_n}a_i^n1_{B_i^n}(X)=g_n(X)$, where
$g_n(x)=\sum_{i=1}^{K_n}a_i^n1_{B^n_i}(x)$. Define $g=\limsup g_n$,
then $g$ is a Borel function. By taking upper limits on both sides
of $W_n=g_n(X)$, we get $W=g(X)$.
\end{proof}

\noindent (ii) Let $X$ be a random variable on a probability space $(\Omega, {\cal F}, {\mathbb P})$, and let $Y$ be a nonnegative random variable on this space. We do not assume that $X$ and $Y$ have a joint density. Nonetheless, show there is a function $g$ such that ${\mathbb E}[Y|X] = g(X)$.

\begin{proof}Note ${\mathbb E}[Y|X]$ is $\sigma(X)$-measurable. By (i),
we can find a Borel function $g$ such that
${\mathbb E}[Y|X]=g(X)$.\end{proof}

\section{Brownian Motion}

\noindent $\blacktriangleright$ {\bf Exercise 3.1.} According to Definition 3.3.3(iii), for $0 \le t < u$, the Brownian motion increment $W(u) - W(t)$ is independent of the $\sigma$-algebra ${\cal F}(t)$. Use this property and property (i) of that definition to show that, for $0 \le t < u_1 < u_2$, the increment $W(u_2) - W(u_1)$ is also indepdent of ${\cal F}_t$.

\begin{proof}
We have ${\cal F}(t)\subset {\cal
F}(u_1)$ and $W(u_2)-W(u_1)$ is independent of ${\cal F}(u_1)$.
So in particular, $W(u_2)-W(u_1)$ is independent of ${\cal F}(t)$.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.2.} Let $W(t)$, $t\ge 0$, be a Brownian motion, and let ${\cal F}(t)$, $t\ge 0$, be a filtration for this Brownian motion. Show that $W^2(t) - t$ is a martingale. (Hint: For $0 \le s \le t$, write $W^2(t)$ as $(W(t)-W(s))^2 + 2W(t) W(s) - W^2(s)$.)

\begin{proof}${\mathbb E}[W^2(t)-W^2(s)|{\cal
F}(s)]={\mathbb E}[(W(t)-W(s))^2+2W(t)W(s)-2W^2(s)|{\cal
F}(s)]=t-s+2W(s){\mathbb E}[W(t)-W(s)|{\cal F}(s)]=t-s$. Simple algebra gives ${\mathbb E}[W^2(t)-t|{\cal F}(s)] = W^2(s) - s$. So $W^2(t)-t$ is a martingale.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.3 (Normal kurtosis).} The {\it kurtosis} of a random variable is defined to be the ratio of its fourth central moment to the square of its variance. For a normal random variable, the kurtosis is $3$. This fact was used to obtain (3.4.7). This exercise verifies this fact.

Let $X$ be a normal random variable with mean $\mu$, so that $X-\mu$ has mean zero. Let the variance of $X$, which is also the variance of $X-\mu$, be $\sigma^2$. In (3.2.13), we computed the moment-generating function of $X-\mu$ to be $\varphi(u)={\mathbb E}e^{u(X-\mu)}=e^{\frac{1}{2}u^2\sigma^2}$, where $u$ is a real variable. Differentiating this function with respect to $u$, we obtain
\[
\varphi'(u) = {\mathbb E} \left[(X-\mu)e^{u(X-\mu)} \right] = \sigma^2 u e^{\frac{1}{2}\sigma^2u^2}
\]
and, in particular, $\varphi'(0) = {\mathbb E}(X-\mu) = 0$. Differentiating again, we obtain
\[
\varphi''(u) = {\mathbb E} \left[(X-\mu)^2 e^{u(X-\mu)}\right] = (\sigma^2 + \sigma^4 u^2) e^{\frac{1}{2}\sigma^2u^2}
\]
and, in particular, $\varphi''(0) = {\mathbb E}[(X-\mu)^2] = \sigma^2$. Differentiate two more times and obtain the normal kurtosis formula ${\mathbb E}[(X-\mu)^4] = 3\sigma^4$.

\begin{solution}
\[
\varphi^{(3)}(u)=2\sigma^4ue^{\frac{1}{2}\sigma^2u^2}+(\sigma^2+\sigma^4u^2)\sigma^2ue^{\frac{1}{2}\sigma^2u^2}=e^{\frac{1}{2}\sigma^2u^2}(3\sigma^4u+\sigma^6u^3),
\]
and
\[
\varphi^{(4)}(u)=\sigma^2ue^{\frac{1}{2}\sigma^2u^2}(3\sigma^4u+\sigma^6u^3)+e^{\frac{1}{2}\sigma^2u^2}(3\sigma^4+3\sigma^6u^2)=e^{\frac{1}{2}\sigma^2u^2}(6\sigma^6u^2+\sigma^8u^4+3\sigma^4).
\]
So ${\mathbb E}[(X-\mu)^4]=\varphi^{(4)}(0)=3\sigma^4$.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.4 (Other variations of Brownian motion).} Theorem 3.4.3 asserts that if $T$ is a positive number and we choose a partition $\Pi$ with points $0 = t_0 < t_1 < t_2 < \cdots < t_n = T$, then as the number $n$ of partition points approaches infinity and the length of the longest subinterval $|\!|\Pi|\!|$ approaches zero, the sample quadratic variation
\[
\sum_{j=0}^{n-1} (W(t_{j+1})-W(t_j))^2
\]
approaches $T$ for almost every path of the Brownian motion $W$. In Remark 3.4.5, we further showed that $\sum_{j=0}^{n-1}(W(t_{j+1})-W(t_j))(t_{j+1}-t_j)$ and $\sum_{j=0}^{n-1}(t_{j+1}-t_j)^2$ have limit zero. We summarize these facts by the multiplication rules
\[
dW(t) dW(t) = dt, dW(t) dt = 0, dtdt = 0. \tag{3.10.1}
\]

\smallskip

\noindent (i) Show that as the number $m$ of partition points approaches infinity and the length of the longest subinterval approaches zero, the sample first variation
\[
\sum_{j=0}^{n-1}|W(t_{j+1})-W(t_j)|
\]
approaches $\infty$ for almost every path of the Brownian motion $W$. (Hint:
\[
\sum_{j=0}^{n-1}(W(t_{j+1})-W(t_j))^2 \le \max_{0\le k \le n-1}|W(t_{k+1}) - W(t_k)| \cdot \sum_{j=0}^{n-1}|W(t_{j+1})-W(t_j)|.)
\]

\begin{proof}Assume there exists $A\in {\cal F}$,
such that ${\mathbb P}(A)>0$ and for every $\omega\in A$,
\[
\limsup_{n}\sum_{j=0}^{n-1}|W_{t_{j+1}}-W_{t_j}|(\omega)<\infty.
\]
Then for every $\omega\in A$,
\begin{eqnarray*}
\sum_{j=0}^{n-1}(W_{t_{j+1}}-W_{t_j})^2(\omega)
&\le& \max_{0\le k\le n-1}|W_{t_{k+1}}-W_{t_k}|(\omega) \cdot \sum_{j=0}^{n-1}|W_{t_{j+1}}-W_{t_j}|(\omega) \\
&\le& \max_{0\le k\le n-1}|W_{t_{k+1}}-W_{t_k}|(\omega) \cdot \limsup_{n}\sum_{j=0}^{n-1}|W_{t_{j+1}}-W_{t_j}|(\omega) \\
&\to& 0,
\end{eqnarray*}
since by uniform continuity of continuous functions over a closed interval, $\lim_{n\to\infty}\max_{0\le k\le
n-1}|W_{t_{k+1}}-W_{t_k}|(\omega)=0$. This is a contradiction with
$\lim_{n\to\infty}\sum_{j=0}^{n-1}(W_{t_{j+1}}-W_{t_j})^2=T$ a.s..
\end{proof}

\noindent (ii) Show that as the number $n$ of partition points approaches infinity and the length of the longest subinterval approaches zero, the sample cubic variation
\[
\sum_{j=0}^{n-1} |W(t_{j+1}) - W(t_j)|^3
\]
approaches zero for almost every path of the Brownian motion $W$.

\begin{proof} We note by an argument similar to (i),
\[
\sum_{j=0}^{n-1}|W(t_{j+1})-W(t_j)|^3\le
\max_{0\le k\le
n-1}|W(t_{k+1})-W(t_k)| \cdot \sum_{j=0}^{n-1}(W(t_{j+1})-W(t_j))^2\to
0
\]
as $n\to\infty$.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.5 (Black-Scholes-Merton formula).} Let the interest rate $r$ and the volatility $\sigma>0$ be constant. Let
\[
S(t) = S(0) e^{\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W(t)}
\]
be a geometric Brownian motion with mean rate of return $r$, where the initial stock price $S(0)$ is positive. Let $K$ be a positive constant. Show that, for $T>0$,
\[
{\mathbb E} \left[e^{-rT} (S(T)-K)^+\right] = S(0) N(d_+(T,S(0))) - Ke^{-rT} N(d_-(T, S(0))),
\]
where
\[
d_{\pm}(T,S(0)) = \frac{1}{\sigma\sqrt{T}} \left[ \log \frac{S(0)}{K} + \left(r\pm \frac{\sigma^2}{2}\right)T \right],
\]
and $N$ is the cumulative standard normal distribution function
\[
N(y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^y e^{-\frac{1}{2}z^2}dz = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-y} e^{-\frac{1}{2}z^2}dz.
\]
\begin{proof}
\begin{eqnarray*}
& & {\mathbb E} \left[e^{-rT}(S_T-K)^+\right] \\
&=& e^{-rT}\int^{\infty}_{\frac{1}{\sigma}\left[\ln\frac{K}{S_0}-(r-\frac{1}{2}\sigma^2)T\right]}
\left(S_0e^{(r-\frac{1}{2}\sigma^2)T+\sigma
x}-K \right)\frac{e^{-\frac{x^2}{2T}}}{\sqrt{2\pi T}}dx\\
&=& e^{-rT}\int^{\infty}_{\frac{1}{\sigma\sqrt{T}}\left[\ln\frac{K}{S_0}-(r-\frac{1}{2}\sigma^2)T\right]}\left(S_0e^{(r-\frac{1}{2}\sigma^2)T+\sigma
\sqrt{T}y}-K\right)\frac{e^{-\frac{y^2}{2}}}{\sqrt{2\pi }}dy\\
&=&S_0e^{-\frac{1}{2}\sigma^2T}\int^{\infty}_{\frac{1}{\sigma\sqrt{T}}\left[\ln\frac{K}{S_0}-(r-\frac{1}{2}\sigma^2)T\right]}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}+\sigma\sqrt{T}y}dy
-Ke^{-rT}\int^{\infty}_{\frac{1}{\sigma\sqrt{T}}\left[\ln\frac{K}{S_0}-(r-\frac{1}{2}\sigma^2)T\right]}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}dy\\
&=&S_0\int^{\infty}_{\frac{1}{\sigma\sqrt{T}}\left[\ln\frac{K}{S_0}-(r-\frac{1}{2}\sigma^2)T\right]-\sigma\sqrt{T}}\frac{1}{\sqrt{2\pi}}e^{-\frac{\xi^2}{2}}d\xi
-Ke^{-rT}N\left(\frac{1}{\sigma\sqrt{T}}\left(\ln\frac{S_0}{K}+(r-\frac{1}{2}\sigma^2)T\right)\right)\\
&=&Ke^{-rT}N(d_+(T,S_0))-Ke^{-rT}N(d_-(T,S_0)).
\end{eqnarray*}
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.6.} Let $W(t)$ be a Brownian motion and let ${\cal F}(t)$, $t\ge 0$, be an associated filtration.

\smallskip

\noindent (i) For $\mu \in {\mathbb R}$, consider the {\it Brownian motion with drift $\mu$}:
\[
X(t) = \mu t + W(t).
\]
Show that for any Borel-measurable function $f(y)$, and for any $0 \le s < t$, the function
\[
g(x) = \frac{1}{\sqrt{2\pi(t-s)}} \int_{-\infty}^{\infty} f(y) \exp \left\{-\frac{(y-x-\mu(t-s))^2}{2(t-s)}\right\}dy
\]
satisfies ${\mathbb E}[f(X(t))|{\cal F}(s)]=g(X(s))$, and hence $X$ has the Markov property. We may rewrite $g(x)$ as $g(x) = \int_{-\infty}^{\infty} f(y) p(\tau, x,y)dy$, where $\tau=t-s$ and
\[
p(\tau,x,y) = \frac{1}{\sqrt{2\pi\tau}} \exp \left\{ -\frac{(y-x-\mu\tau)^2}{2\tau} \right\}
\]
is the {\it transition density} for Brownian motion with drift $\mu$.

\begin{proof}
\begin{eqnarray*}
{\mathbb E}[f(X_t)|{\cal
F}_t]
&=& {\mathbb E}[f(W_t-W_s+a)|{\cal F}_s]|_{a=W_s+\mu t} \\
&=& {\mathbb E}[f(W_{t-s}+a)]|_{a=W_s+\mu t}\\
&=&\int_{-\infty}^{\infty}f(x+W_s+\mu
t)\frac{e^{-\frac{x^2}{2(t-s)}}}{\sqrt{2\pi
(t-s)}}dx\\
&=&\int_{-\infty}^{\infty}f(y)\frac{e^{-\frac{(y-W_s-\mu s-\mu
(t-s))^2}{2(t-s)}}}{\sqrt{2\pi (t-s)}}dy\\
&=&g(X_s).
\end{eqnarray*}
 So ${\mathbb E}[f(X_t)|{\cal
F}_s]=\int_{-\infty}^{\infty}f(y)p(t-s,X_s,y)dy$ with
$p(\tau,x,y)=\frac{1}{\sqrt{2\pi
\tau}}e^{-\frac{(y-x-\mu\tau)^2}{2\tau}}$.
\end{proof}

\noindent (ii) For $\nu \in {\mathbb R}$ and $\sigma > 0$, consider the {\it geometric Brownian motion}
\[
S(t) = S(0) e^{\sigma W(t) + \nu t}.
\]
Set $\tau = t-s$ and
\[
p(\tau, x, y) = \frac{1}{\sigma y \sqrt{2\pi \tau}} \exp \left\{ -\frac{\left(\log\frac{y}{x}-\nu\tau\right)^2}{2\sigma^2\tau} \right\}.
\]
Show that for any Borel-measurable function $f(y)$ and for any $0 \le s < t$ the function
\[
g(x) = \int_0^{\infty} f(y) p(\tau, x, y) dy\footnote{The textbook wrote $\int_0^{\infty}h(y)p(\tau,x,y)dy$ by mistake.}
\]
satisfies ${\mathbb E}[f(S(t))|{\cal F}(s)] = g(S(s))$ and hence $S$ has the Markov property and $p(\tau,x,y)$ is its transition density.

\begin{proof} ${\mathbb E}[f(S_t)|{\cal F}_s]={\mathbb E}[f(S_0e^{\sigma X_t})|{\cal
F}_s]$ with $\mu=\frac{v}{\sigma}$. So by (i),
\begin{eqnarray*}
{\mathbb E}[f(S_t)|{\cal F}_s] &=&\int_{-\infty}^{\infty}f(S_0e^{\sigma
y})\frac{1}{\sqrt{2\pi(t-s)}}e^{-\frac{[y-X_s-\mu(t-s)]^2}{2(t-s)}}dy\\
&\stackrel{S_0e^{\sigma
y}=z}{=}&\int_0^{\infty}f(z)\frac{1}{\sqrt{2\pi
(t-s)}}e^{-\frac{\left[\frac{1}{\sigma}\ln\frac{z}{S_0}-\frac{1}{\sigma}\ln\frac{S_s}{S_0}-\mu(t-s)\right]^2}{2}}\frac{dz}{\sigma
z}\\
&=&\int_0^{\infty}f(z)\frac{e^{-\frac{\left[\ln\frac{z}{S_s}-v(t-s)\right]^2}{2\sigma^2(t-s)}}}{\sigma
z\sqrt{2\pi (t-s)}}dz\\
&=&\int_0^{\infty}f(z)p(t-s,S_s,z)dz\\
&=&g(S_s).
\end{eqnarray*}
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.7.}  Theorem 3.6.2 provides the Laplace transform of the density of the first passage time for Brownian motion. This problem derives the analogous formula for Brownian motions with drift.  Let $W$ be a Brownian motion. Fix $m>0$ and $\mu \in {\mathbb R}$. For $0 \le t < \infty$, define
\begin{eqnarray*}
X(t) &=& \mu t + W(t), \\
\tau_m = \min \{ t\ge 0; X(t) = m \}.
\end{eqnarray*}
As usual, we set $\tau_m = \infty$ if $X(t)$ never reaches the level $m$. Let $\sigma$ be a positive number and set
\[
Z(t) = \exp \left\{\sigma X(t) - \left(\sigma \mu + \frac{1}{2}\sigma^2\right) t \right\}.
\]

\smallskip

\noindent (i) Show that $Z(t)$, $t\ge 0$, is a martingale.

\begin{proof}
\begin{eqnarray*}
& & {\mathbb E}\left[\left.\frac{Z_t}{Z_s}\right|{\cal F}_s\right] \\
&=& {\mathbb E}\left[\exp\left\{\sigma(W_t-W_s)+\sigma\mu(t-s)-\left(\sigma\mu+\frac{\sigma^2}{2}\right)(t-s)\right\}\right] \\
&=& \exp\left\{\sigma\mu(t-s)-\left(\sigma\mu+\frac{\sigma^2}{2}\right)(t-s)\right\} \cdot \int_{-\infty}^{\infty} e^{\sigma\sqrt{t-s} \cdot x}\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}dx\\
&=& \exp\left\{-\frac{\sigma^2}{2}(t-s)\right\} \cdot \int_{-\infty}^{\infty} \frac{e^{-\frac{(x-\sigma\sqrt{t-s})^2}{2}}}{\sqrt{2\pi}}dx \cdot \exp\left\{\frac{\sigma^2}{2}(t-s)\right\} \\
&=& 1.
\end{eqnarray*}
\end{proof}

\noindent (ii) Use (i) to conclude that
\[
{\mathbb E} \left[\exp \left\{\sigma X(t\wedge\tau_m) - \left(\sigma\mu + \frac{1}{2}\sigma^2\right) (t\wedge\tau_m) \right\} \right] = 1, \; t\ge 0.
\]

\begin{proof}By optional stopping theorem,
${\mathbb E}[Z_{t\wedge\tau_m}]={\mathbb E}[Z_0]=1$, that is,
\[
{\mathbb E}\left[\exp\left\{\sigma
X_{t\wedge\tau_m}-\left(\sigma\mu+\frac{\sigma^2}{2}\right)t\wedge\tau_m\right\}\right]=1.
\]
\end{proof}

\noindent (iii) Now suppose $\mu \ge 0$. Show that, for $\sigma > 0$,
\[
{\mathbb E} \left[\exp\left\{\sigma m - \left(\sigma\mu+\frac{1}{2}\sigma^2\right)\tau_m \right\} 1_{\{\tau_m < \infty\}} \right] = 1.
\]

\begin{proof} If $\mu\ge 0$ and $\sigma>0$,
$Z_{t\wedge\tau_m}\le e^{\sigma m}$. By bounded convergence theorem,
\[
{\mathbb E}[1_{\{\tau_m<\infty\}}Z_{\tau_m}]={\mathbb E}[\lim_{t\to\infty}Z_{t\wedge
\tau_m}]=\lim_{t\to\infty}{\mathbb E}[Z_{t\wedge\tau_m}]=1,\]
 since on the
event $\{\tau_m=\infty\}$, $Z_{t\wedge\tau_m}\le e^{\sigma
m-\frac{1}{2}\sigma^2t}\to 0$ as $t\to\infty$. Therefore,
${\mathbb E}\left[e^{\sigma m-(\sigma\mu+\frac{\sigma^2}{2})\tau_m}1_{\{\tau_m<\infty\}}\right]=1$. Let
$\sigma\downarrow 0$, by bounded convergence theorem, we have
${\mathbb P}(\tau_m<\infty)=1$. Let $\sigma\mu+\frac{\sigma^2}{2}=\alpha$, we
get
\[
{\mathbb E}[e^{-\alpha\tau_m}]=e^{-\sigma m}=e^{m\mu-m\sqrt{2\alpha+\mu^2}}.
\]
\end{proof}

\noindent (iv) Show that if $\mu > 0$, then ${\mathbb E}\tau_m < \infty$. Obtain a formula for ${\mathbb E} \tau_m$. (Hint: Differentiate the formula in (iii) with respect to $\alpha$.)

\begin{proof}We note for $\alpha>0$,
${\mathbb E}[\tau_me^{-\alpha\tau_m}]<\infty$ since $xe^{-\alpha x}$ is
bounded on $[0,\infty)$. So by an argument similar to Exercise 1.8,
${\mathbb E}[e^{-\alpha\tau_m}]$ is differentiable and
\[
\frac{\partial}{\partial
\alpha}{\mathbb E}[e^{-\alpha\tau_m}]=-{\mathbb E}[\tau_me^{-\alpha\tau_m}]=e^{m\mu-m\sqrt{2\alpha+\mu^2}}\frac{-m}{\sqrt{2\alpha+\mu^2}}.
\]
Let $\alpha\downarrow 0$, by monotone increasing theorem,
${\mathbb E}[\tau_m]=\frac{m}{\mu}<\infty$ for $\mu>0$.
\end{proof}

\noindent (v) Now suppose $\mu < 0$. Show that, for $\sigma > -2\mu$,
\[
{\mathbb E} \left[\exp \left\{ \sigma m - \left(\sigma \mu + \frac{1}{2}\sigma^2\right)\tau_m \right\} 1_{\{\tau_m < \infty\}}\right] = 1.
\]
Use this fact to show that ${\mathbb P}\{\tau_m < \infty\} = e^{-2m|\mu|}$,\footnote{The textbook wrote $e^{-2x|\mu|}$ by mistake.} which is strictly less than one, and to obtain the Laplace transform
\[
{\mathbb E} e^{-\alpha\tau_m} = e^{m \mu - m \sqrt{2\alpha + \mu^2}} \; \mbox{for all $\alpha>0$.}
\]

\begin{proof} By $\sigma>-2\mu>0$, we get
$\sigma\mu+\frac{\sigma^2}{2}>0$. Then $Z_{t\wedge\tau_m}\le
e^{\sigma m}$ and on the event $\{\tau_m=\infty\}$,
$Z_{t\wedge\tau_m}\le e^{\sigma
m-(\frac{\sigma^2}{2}+\sigma\mu)t}\to 0$ as $t\to\infty$. Therefore,
\[
{\mathbb E}\left[e^{\sigma
m-(\sigma\mu+\frac{\sigma^2}{2})\tau_m}1_{\{\tau_m<\infty\}}\right]={\mathbb E}[\lim_{t\to\infty}Z_{t\wedge\tau_m}]=\lim_{t\to\infty}{\mathbb E}[Z_{t\wedge\tau_m}]=1.
\]
Let $\sigma\downarrow -2\mu$, we then get ${\mathbb P}(\tau_m<\infty)=e^{2\mu
m}=e^{-2|\mu|m}<1$. Set $\alpha=\sigma\mu+\frac{\sigma^2}{2} > 0$ and we
have
\[
{\mathbb E}[e^{-\alpha\tau_m}]={\mathbb E}[e^{-\alpha\tau_m}1_{\{\tau_m<\infty\}}]=e^{-\sigma
m}=e^{m\mu-m\sqrt{2\alpha+\mu^2}}.
\]
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.8.} This problem presents the convergence of the distribution of stock prices in a sequence of binomial models to the distribution of geometric Brownian motion. In contrast to the analysis of Subsection 3.2.7, here we allow the interest rate to be different from zero.

Let $\sigma > 0$ and $r \ge 0$ be given. For each positive integer $n$, we consider a binomial model taking $n$ steps per unit time. In this model, the interest rate per period is $\frac{r}{n}$, the up factor is $u_n = e^{\sigma/\sqrt{n}}$, and the down factor is $d_n = e^{-\sigma/\sqrt{n}}$. The risk-neutral probabilities are then
\[
\widetilde p_n = \frac{\frac{r}{n}+1-e^{-\sigma/\sqrt{n}}}{e^{\sigma/\sqrt{n}}-e^{-\sigma/\sqrt{n}}}, \;
\widetilde q_n = \frac{e^{\sigma/\sqrt{n}}-\frac{r}{n}-1}{e^{\sigma/\sqrt{n}} - e^{-\sigma/\sqrt{n}}}.
\]
Let $t$ be an arbitrary positive rational number, and for each positive integer $n$ for which $nt$ is an integer, define
\[
M_{nt,n} = \sum_{k=1}^{nt} X_{k,n},
\]
where $X_{1,n}$, $\cdots$, $X_{nt,n}$\footnote{The textbook wrote $X_{n,n}$ by mistake.} are independent, identically distributed random variables with
\[
\widetilde {\mathbb P}\{X_{k,n}=1\}=\widetilde p_n, \; \widetilde {\mathbb P}\{X_{k,n}=-1\} = \widetilde q_n, k= 1, \cdots, nt\footnote{The textbook wrote $n$ by mistake.}.
\]
The stock price at time $t$ in this binomial model, which is the result of $nt$ steps from the initial time, is given by (see (3.2.15) for a similar equation)
\begin{eqnarray*}
S_n(t) &=& S(0) u_n^{\frac{1}{2}(nt + M_{nt,n})} d_n^{\frac{1}{2}(nt-M_{nt,n})} \\
&=& S(0) \exp \left\{\frac{\sigma}{2\sqrt{n}}(nt + M_{nt,n})\right\}\footnote{The textbook wrote $M_{nT,n}$ by mistake.} \exp \left\{-\frac{\sigma}{2\sqrt{n}} (nt-M_{nt,n})\right\} \\
&=& S(0) \exp \left\{\frac{\sigma}{\sqrt{n}}M_{nt,n}\right\}.
\end{eqnarray*}
This problem shows that as $n\to\infty$, the distribution of the sequence of random variables $\frac{\sigma}{\sqrt{n}}M_{nt,n}$ appearing in the exponent above converges to the normal distribution with mean $\left(r-\frac{1}{2}\sigma^2\right)t$ and variance $\sigma^2t$. Therefore, the limiting distribution of $S_n(t)$ is the same as the distribution of the geometric Brownian motion $S(0) \exp \{\sigma W(t) + (r-\frac{1}{2}\sigma)t\}$ at time $t$.

\smallskip

\noindent (i) Show that the moment-generating function $\varphi_n(u)$ of $\frac{1}{\sqrt{n}}M_{nt,n}$ is given by
\[
\varphi_n(u) = \left[e^{\frac{u}{\sqrt{n}}}\left(\frac{\frac{r}{n}+1-e^{-\sigma/\sqrt{n}}}{e^{\sigma/\sqrt{n}}-e^{-\sigma/\sqrt{n}}}\right) - e^{-\frac{u}{\sqrt{n}}} \left(\frac{\frac{r}{n}+1-e^{\sigma/\sqrt{n}}}{e^{\sigma/\sqrt{n}}-e^{-\sigma/\sqrt{n}}}\right)\right]^{nt}.
\]

\begin{proof}
\begin{eqnarray*}
\varphi_n(u) &=&\widetilde {\mathbb E}\left[e^{u\frac{1}{\sqrt{n}}M_{nt,n}}\right]
=\left(\widetilde {\mathbb E}\left[e^{\frac{u}{\sqrt{n}}X_{1,n}}\right]\right)^{nt}
=(e^{\frac{u}{\sqrt{n}}}\widetilde
p_n+e^{-\frac{u}{\sqrt{n}}}\widetilde q_n)^{nt}\\
&=&\left[
e^{\frac{u}{\sqrt{n}}}\left(\frac{\frac{r}{n}+1-e^{-\frac{\sigma}{\sqrt{n}}}}{e^{\frac{\sigma}{\sqrt{n}}}-e^{-\frac{\sigma}{\sqrt{n}}}}\right)+
e^{-\frac{u}{\sqrt{n}}}\left(\frac{-\frac{r}{n}-1+e^{\frac{\sigma}{\sqrt{n}}}}{e^{\frac{\sigma}{\sqrt{n}}}-e^{-\frac{\sigma}{\sqrt{n}}}}\right)
 \right]^{nt}.
\end{eqnarray*}
\end{proof}

\noindent (ii) We want to compute
\[
\lim_{n\to\infty}\varphi_n(u) = \lim_{x\downarrow 0} \varphi_{\frac{1}{x^2}}(u),
\]
where we have made the change of variable $x=\frac{1}{\sqrt{n}}$. To do this, we will compute $\log \varphi_{\frac{1}{x^2}} (u)$ and then take the limit as $x\downarrow 0$. Show that
\[
\log \varphi_{\frac{1}{x^2}}(u) = \frac{t}{x^2} \log \left[\frac{(rx^2+1)\sinh ux + \sinh (\sigma-u)x}{\sinh \sigma x}\right]
\]
(the definitions are $\sinh z = \frac{e^z-e^{-z}}{2}$, $\cosh z = \frac{e^z + e^{-z}}{2}$), and use the formula
\[
\sinh(A-B) = \sinh A \cosh B - \cosh A \sinh B
\]
to rewrite this as
\[
\log \varphi_{\frac{1}{x^2}}(u) = \frac{t}{x^2} \log \left[\cosh ux + \frac{(rx^2+1-\cosh\sigma x)\sinh ux}{\sinh \sigma x}\right].
\]

\begin{proof}
\[
\varphi_{\frac{1}{x^2}}(u)=\left[e^{ux}\left(\frac{rx^2+1-e^{-\sigma
x}}{e^{\sigma x}-e^{-\sigma
x}}\right)-e^{-ux}\left(\frac{rx^2+1-e^{\sigma x}}{e^{\sigma
x}-e^{-\sigma x}}\right)\right]^{\frac{t}{x^2}}.
\]So,
\begin{eqnarray*}
\log\varphi_{\frac{1}{x^2}}(u)&=&\frac{t}{x^2}\log\left[\frac{(rx^2+1)(e^{ux}-e^{-ux})+e^{(\sigma-u)x}-e^{-(\sigma
-u)x}}{e^{\sigma x}-e^{-\sigma x}}\right]\\
&=&\frac{t}{x^2}\log\left[\frac{(rx^2+1)\sinh u x+\sinh (\sigma-u)x}{\sinh \sigma x}\right]\\
&=&\frac{t}{x^2}\log\left[\frac{(rx^2+1)\sinh u x+\sinh \sigma x\cosh
u x-\cosh \sigma x\sinh u x}{\sinh \sigma x}\right]\\
&=&\frac{t}{x^2}\log\left[\cosh u x+\frac{(rx^2+1-\cosh \sigma
x)\sinh u x}{\sinh \sigma x}\right].
\end{eqnarray*}
\end{proof}

\noindent (iii) Use the Taylor series expansions
\[
\cosh z = 1 + \frac{1}{2} z^2 + O(z^4), \sinh z = z + O(z^3),
\]
to show that
\[
\cosh ux + \frac{(rx^2+1-\cosh\sigma x)\sinh ux}{\sinh \sigma x} = 1+\frac{1}{2}u^2 x^2 + \frac{rux^2}{\sigma} - \frac{1}{2}ux^2\sigma + O(x^4). \tag{3.10.2}
\]
The notation $O(x^j)$ is used to represent terms of the order $x^j$.

\begin{proof}
\begin{eqnarray*}
& &\cosh u x+\frac{(rx^2+1-\cosh \sigma x)\sinh u x}{\sinh \sigma x}\\
&=&1+\frac{u^2x^2}{2}+O(x^4)+\frac{(rx^2+1-1-\frac{\sigma^2x^2}{2}+O(x^4))(ux+O(x^3))}{\sigma
x+O(x^3)}\\
&=&1+\frac{u^2x^2}{2}+\frac{(r-\frac{\sigma^2}{2})ux^3+O(x^5)}{\sigma x+O(x^3)}+O(x^4)\\
&=&1+\frac{u^2x^2}{2}+\frac{(r-\frac{\sigma^2}{2})ux^3(1+O(x^2))}{\sigma x(1+O(x^2))}+O(x^4)\\
&=&1+\frac{u^2x^2}{2}+\frac{rux^2}{\sigma}-\frac{1}{2}\sigma
ux^2+O(x^4).
\end{eqnarray*}
\end{proof}

\noindent (iv) Use the Taylor series expansion $\log(1+x)=x+O(x^2)$ to compute $\lim_{x\downarrow 0} \log \varphi_{\frac{1}{x^2}}(u)$. Now explain how you know that the limiting distribution for $\frac{\sigma}{\sqrt{n}}M_{nt,n}$ is normal with mean $\left(r-\frac{1}{2}\sigma^2\right)t$ and variance $\sigma^2t$.
\begin{solution}
\[
\log
\varphi_{\frac{1}{x^2}}=\frac{t}{x^2}\log\left(1+\frac{u^2x^2}{2}+\frac{ru}{\sigma}x^2-\frac{\sigma
u
x^2}{2}+O(x^4)\right)=\frac{t}{x^2}\left(\frac{u^2x^2}{2}+\frac{ru}{\sigma}x^2-\frac{\sigma
u x^2}{2}+O(x^4)\right).
\]
So $\lim_{x\downarrow
0}\log\varphi_{\frac{1}{x^2}}(u)=t(\frac{u^2}{2}+\frac{ru}{\sigma}-\frac{\sigma
u}{2})$, and $\widetilde
{\mathbb E}\left[e^{u\frac{1}{\sqrt{n}}M_{nt,n}}\right]=\varphi_n(u)\to\frac{1}{2}tu^2+t(\frac{r}{\sigma}-\frac{\sigma}{2})u$.
By the one-to-one correspondence between distribution and moment
generating function\footnote{This correspondence holds only under certain conditions, which normal distribution certainly satisfies. For details, see Shiryaev \cite[page~293-296]{Shiryaev95}.}, $(\frac{1}{\sqrt{n}}M_{nt,n})_n$ converges to a
Gaussian random variable with mean
$t(\frac{r}{\sigma}-\frac{\sigma}{2})$ and variance $t$. Hence
$(\frac{\sigma}{\sqrt{n}}M_{nt,n})_n$ converges to a Gaussian random
variable with mean $t(r-\frac{\sigma^2}{2})$ and variance
$\sigma^2t$.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.9 (Laplace transform of first passage density).} The solution to this problem is long and technical. It is included for the sake of completeness, but the reader may safely skip it.

Let $m>0$ be given, and define
\[
f(t,m) = \frac{m}{t\sqrt{2\pi t}} \exp \left\{-\frac{m^2}{2t}\right\}.
\]
According to (3.7.3) in Theorem 3.7.1, $f(t,m)$ is the density in the variable $t$ of the first passage time $\tau_m = \min \{t\ge 0; W(t) = m\}$, where $W$ is a Brownian motion without drift. Let
\[
g(\alpha, m) = \int_0^{\infty} e^{-\alpha t} f(t,m) dt, \alpha > 0,
\]
be the Laplace transform of the density $f(t,m)$. This problem verifies that $g(\alpha, m) = e^{-m\sqrt{2\alpha}}$, which is the formula derived in Theorem 3.6.2.

\smallskip

\noindent (i) For $k\ge 1$, define
\[
\alpha_k(m) = \frac{1}{\sqrt{2\pi}} \int_0^{\infty} t^{-k/2} \exp \left\{-\alpha t - \frac{m^2}{2t}\right\} dt,
\]
so $g(\alpha,m)=ma_3(m)$. Show that
\begin{eqnarray*}
g_m(\alpha,m) &=& a_3(m) - m^2 a_5(m), \\
g_{mm}(\alpha,m) &=& -3ma_5(m) + m^3a_7(m).
\end{eqnarray*}

\begin{proof} We note
\begin{eqnarray*}
\frac{\partial}{\partial m}a_k(m)
&=& \frac{1}{\sqrt{2\pi}} \int_0^{\infty} t^{-k/2}\frac{\partial}{\partial m}\exp\left\{-\alpha t - \frac{m^2}{2t} \right\}dt \\
&=& \frac{1}{\sqrt{2\pi}} \int_0^{\infty} t^{-k/2}\exp\left\{-\alpha t - \frac{m^2}{2t} \right\} \left(-\frac{m}{t}\right) dt \\
&=& -m a_{k+2}(m).
\end{eqnarray*}
So
\[
g_m(\alpha,m) = \frac{\partial}{\partial m} [m a_3(m)] = a_3(m) - m^2 a_5(m),
\]
and
\[
g_{mm}(\alpha,m) = -m a_5(m)-2m a_5(m) + m^3 a_7(m) = -3m a_5(m) + m^3 a_7(m).
\]
\end{proof}

\noindent (ii) Use integration by parts to show that
\[
a_5(m) = -\frac{2\alpha}{3} a_3(m) + \frac{m^2}{3} a_7(m).
\]

\begin{proof}
For $k>2$ and $\alpha >0$, we have
\begin{eqnarray*}
a_k(m) &=& \frac{1}{\sqrt{2\pi}} \int_0^{\infty} t^{-k/2}\exp\left\{-\alpha t - \frac{m^2}{2t}\right\}dt \\
&=& -\frac{2}{k-2} \cdot \frac{1}{\sqrt{2\pi}} \int_0^{\infty} \exp\left\{-\alpha t - \frac{m^2}{2t}\right\}dt^{-(k-2)/2} \\
&=& -\frac{2}{k-2} \cdot \frac{1}{\sqrt{2\pi}}  \left[\left. \exp\left\{-\alpha t - \frac{m^2}{2t}\right\} t^{-(k-2)/2}\right|_{0}^{\infty} \right. \\
& & \left.- \int_0^{\infty} t^{-(k-2)/2}\exp\left\{-\alpha t - \frac{m^2}{2t}\right\}\left(-\alpha+\frac{m^2}{2t^2}\right)dt\right] \\
&=& -\frac{2}{k-2} \cdot \frac{1}{\sqrt{2\pi}}  \left[\alpha \int_0^{\infty} t^{-(k-2)/2}\exp\left\{-\alpha t - \frac{m^2}{2t}\right\}dt - \frac{m^2}{2} \int_0^{\infty} t^{-(k+2)/2}\exp\left\{-\alpha t - \frac{m^2}{2t}\right\}dt\right] \\
&=& -\frac{2\alpha}{k-2}a_{k-2}(m) + \frac{m^2}{k-2} a_{k+2}(m).
\end{eqnarray*}
Plug in $k=5$, we obtain $a_5(m) = -\frac{2\alpha}{3}a_3(m) + \frac{m^2}{3}a_7(m)$.
\end{proof}

\noindent (iii) Use (i) and (ii) to show that $g$ satisfies the second-order ordinary differential equation
\[
g_{mm}(\alpha,m)=2\alpha g(\alpha,m).
\]

\begin{proof}
We note
\begin{eqnarray*}
g_{mm}(\alpha, m) = -3ma_5(m) + m^3a_7(m) = 2\alpha m a_3(m) - m^3 a_7(m) + m^3 a_7(m) = 2\alpha m a_3(m) = 2\alpha g(\alpha,m).
\end{eqnarray*}
\end{proof}

\noindent (iv) The general solution to a second-order ordinary differential equation of the form
\[
ay''(m) + by'(m) + cy(m) = 0
\]
is
\[
y(m) = A_1 e^{\lambda_1 m} + A_2 e^{\lambda_2 m},
\]
where $\lambda_1$ and $\lambda_2$ are roots of the {\it characteristic equation}
\[
a\lambda^2 + b\lambda + c = 0.
\]
Here we are assuming that these roots are distinct. Find the general solution of the equation in (iii) when $\alpha > 0$. This solution has two undetermined parameters $A_1$ and $A_2$, and these may depend on $\alpha$.

\begin{solution}
The characteristic equation of $g_{mm}(\alpha, m) = 2\alpha g(\alpha,m)$ is
\[
\lambda^2 - 2\alpha = 0.
\]
So $\lambda_1 = \sqrt{2\alpha}$ and $\lambda_2 = -\sqrt{2\alpha}$, and the general solution of the equation in (iii) is
\[
A_1 e^{\sqrt{2\alpha} m} + A_2 e^{-\sqrt{2\alpha} m}.
\]
\end{solution}

\noindent (v) Derive the bound
\[
g(\alpha, m) \le \frac{m}{\sqrt{2\pi}} \int_0^m \sqrt{\frac{m}{t}} t^{-3/2} \exp \left\{-\frac{m^2}{2t}\right\} dt + \frac{1}{\sqrt{2\pi m}} \int_m^{\infty} e^{-\alpha t} dt
\]
and use it to show that, for every $\alpha >0$,
\[
\lim_{m\to\infty} g(\alpha, m) = 0.
\]
Use this fact to determine one of the parameters in the general solution to the equation in (iii).

\begin{solution} We have
\begin{eqnarray*}
g(\alpha, m) &=& m a_3(m) \\
&=& \frac{m}{\sqrt{2\pi}} \int_0^{\infty} t^{-3/2} \exp \left\{-\alpha t - \frac{m^2}{2t} \right\} dt \\
&=& \frac{m}{\sqrt{2\pi}} \left(\int_0^{m} t^{-3/2} \exp \left\{-\alpha t - \frac{m^2}{2t} \right\} dt + \int_m^{\infty} t^{-3/2} \exp \left\{-\alpha t - \frac{m^2}{2t} \right\} dt \right) \\
&\le& \frac{m}{\sqrt{2\pi}} \left(\int_0^{m} t^{-3/2} \exp \left\{-\alpha t - \frac{m^2}{2t} \right\} dt + \int_m^{\infty} m^{-3/2} \exp \left\{-\alpha t \right\} dt \right) \\
&=& \frac{m}{\sqrt{2\pi}} \int_0^m e^{-\alpha t} t^{-3/2} \exp \left\{-\frac{m^2}{2t}\right\} dt + \frac{1}{\sqrt{2\pi m}} \int_m^{\infty} e^{-\alpha t} dt.
\end{eqnarray*}
Over the interval $(0, m]$, $e^{-\alpha t}$ is monotone decreasing with a range of $[e^{-\alpha m}, 1)$ and $\sqrt{\frac{m}{t}}$ is monotone decreasing with a range of $[1,\infty)$. so $e^{-\alpha t} < \sqrt{\frac{m}{t}}$ over the interval $(0, m]$. This gives the desired bound
\[
g(\alpha, m) \le \frac{m}{\sqrt{2\pi}} \int_0^m \sqrt{\frac{m}{t}} t^{-3/2} \exp \left\{-\frac{m^2}{2t}\right\} dt + \frac{1}{\sqrt{2\pi m}} \int_m^{\infty} e^{-\alpha t} dt.
\]

Now, let $m\to\infty$. Obviously $\lim_{m\to\infty} \frac{1}{\sqrt{2\pi m}} \int_m^{\infty} e^{-\alpha t} dt = 0$, while the first term through the change of variable $-\frac{1}{t}=u$ satisfies
\begin{eqnarray*}
\frac{m}{\sqrt{2\pi}} \int_0^m \sqrt{\frac{m}{t}} t^{-3/2} \exp \left\{-\frac{m^2}{2t}\right\} dt
&=& \frac{m^{3/2}}{\sqrt{2\pi}} \int_0^m t^{-2} \exp \left\{-\frac{m^2}{2t}\right\} dt \\
&=& \frac{m^{3/2}}{\sqrt{2\pi}} \int_{-\infty}^{-\frac{1}{m}} e^{\frac{m^2}{2} u}du \\
&=& \frac{m^{3/2}}{\sqrt{2\pi}}  \cdot \frac{e^{-m/2}}{m^2/2} \to 0
\end{eqnarray*}
as $m\to\infty$. Combined, we can conclude that for every $\alpha > 0$, $\lim_{m\to\infty} g(\alpha, m)= \lim_{m\to\infty} (A_1 e^{\sqrt{2\alpha} m} + A_2 e^{-\sqrt{2\alpha} m}) = 0$, which requires $A_1 = 0$ as a necessary condition. Therefore, $g(\alpha, m) = A_2 e^{-\sqrt{2\alpha} m}$.

\end{solution}

\noindent (vi) Using first the change of variable $s= t/m^2$ and then the change of variable $y=1/\sqrt{s}$, show that
\[
\lim_{m\downarrow 0} g(\alpha, m) = 1.
\]
Using this fact to determine the other parameter in the general solution to the equation in (iii).

\begin{solution} Following the hint of the problem, we have
\begin{eqnarray*}
g(\alpha, m) &=& \frac{m}{\sqrt{2\pi}} \int_0^{\infty} t^{-3/2} e^{-\alpha t - \frac{m^2}{2t}} dt \\
&\overset{s=t/m^2}{=}& \frac{m}{\sqrt{2\pi}} \int_0^{\infty} (sm^2)^{-3/2} e^{-\alpha m^2s - \frac{1}{2s}} m^2 ds \\
&=& \frac{1}{\sqrt{2\pi}} \int_0^{\infty} s^{-3/2} e^{-\alpha m^2s - \frac{1}{2s}} ds \\
&\overset{y=1/\sqrt{s}}{=}&  \frac{1}{\sqrt{2\pi}} \int_0^{\infty} y^3 e^{-\alpha m^2\frac{1}{y^2} - \frac{y^2}{2}}\frac{2}{y^3} dy \\
&=&  \frac{2}{\sqrt{2\pi}} \int_0^{\infty}  e^{-\alpha m^2\frac{1}{y^2} - \frac{y^2}{2}} dy.
\end{eqnarray*}
So by dominated convergence theorem,
\[
\lim_{m\downarrow 0} g(\alpha, m) = \frac{2}{\sqrt{2\pi}} \int_0^{\infty}  \lim_{m\downarrow 0} e^{-\alpha m^2\frac{1}{y^2} - \frac{y^2}{2}} dy  = \frac{2}{\sqrt{2\pi}} \int_0^{\infty}  e^{- \frac{y^2}{2}} dy = 1.
\]
We already proved in (v) that $g(\alpha, m) = A_2 e^{-\sqrt{2\alpha} m}$. So we must have $A_2=1$ and hence $g(\alpha, m) = e^{-\sqrt{2\alpha} m}$.
\end{solution}



\section{Stochastic Calculus}

$\bigstar$ {\bf Comments}:

\medskip

1) To see how we obtain the solution to the Vasicek interest rate model (equation (4.4.33)), we recall the method of {\it integrating factors}, a technique often used in solving first-order linear ordinary differential equations (see, for example, Logan \cite[page~74]{Logan10}). Starting from (4.4.32)
\[
dR(t) = (\alpha - \beta R(t)) dt + \sigma dW(t),
\]
we move the term containing $R(t)$ to the left side of the equation and multiple both sides by the integrating factor $e^{\beta t}$:
\[
e^{\beta t}[R(t)d(\beta t) + dR(t)] = e^{\beta t} [ \alpha dt + \sigma dW(t)].
\]
Applying It\^{o}'s formula lets us know the left side of the equation is
\[
d(e^{\beta t}R(t)).
\]
Integrating from $0$ to $t$ on both sides, we obtain
\[
e^{\beta t} R(t) - R(0) = \frac{\alpha}{\beta} (e^{\beta t} - 1) + \sigma \int_0^t e^{\beta s}dW(s).
\]
This gives us (4.4.33)
\[
R(t) = e^{-\beta t}R(0) + \frac{\alpha}{\beta} (1-e^{-\beta t}) + \sigma e^{-\beta t} \int_0^t e^{\beta s} dW(s).
\]

\medskip

2) The distribution of CIR interest rate process $R(t)$ for each positive $t$ can be found in reference [41] of the textbook (Cox, J. C., INGERSOLL, J. E., AND ROSS, S. (1985) A theory of the term structure of interest rates, {\it Econometrica} {\bf 53}, 373-384).

\medskip

3) Regarding to the comment at the beginning of Section 4.5.2, ``Black, Scholes, and Merton argued that the value of this call at any time should depend on the time (more precisely, on the time to expiration) and on the value of the stock price at that time", we note it is justified by risk-neutral pricing and Markov property:
\[
C(t) = \widetilde {\mathbb E}[e^{-r(T-t)}(S_T-K)^+|{\cal F}(t)] = \widetilde {\mathbb E}[e^{-r(T-t)}(S_T-K)^+|S_t] = c(t,S_t).
\]
For details, see Chapter 5.

\medskip

4) For the PDE approach to solving the Black-Scholes-Merton equation, see Wilmott \cite{Wilmott95} for details.

\bigskip

\noindent $\blacktriangleright$ {\bf Exercise 4.1.} Suppose $M(t)$, $0 \le t \le T$, is a martingale with respect to some filtration ${\cal F}(t)$, $0 \le t \le T$. Let $\Delta(t)$, $0 \le t \le T$, be a simple process adapted to ${\cal F}(t)$ (i.e., there is a partition $\Pi = \{t_0, t_1, \cdots, t_n\}$ of $[0, T]$ such that, for every $j$, $\Delta(t_j)$ is ${\cal F}(t_j)$-measurable and $\Delta(t)$ is constant in $t$ on each subinterval $[t_j, t_{j+1}))$. For $t\in [t_k, t_{k+1})$, define the stochastic integral
\[
I(t) = \sum_{j=0}^{k-1} \Delta(t_j) [M(t_{j+1}) - M(t_j)] + \Delta(t_k) [M(t) - M(t_k)].
\]
We think of $M(t)$ as the price of an asset at time $t$ and $\Delta(t_j)$ as the number of shares of the asset held by an investor between times $t_j$ and $t_{j+1}$. Then $I(t)$ is the capital gains that accrue to the investor between times $0$ and $t$. Show that $I(t)$, $0 \le t \le T$, is a martingale.

\begin{proof}Fix $t$ and for any $s<t$, we assume
$s\in [t_m,t_{m+1})$ for some $m$.

{\it Case 1}. $m=k$. Then
$I(t)-I(s)=\Delta_{t_k}(M_t-M_{t_k})-\Delta_{t_k}(M_s-M_{t_k})=\Delta_{t_k}(M_t-M_s)$.
So ${\mathbb E}[I(t)-I(s)|{\cal F}_t]=\Delta_{t_k}{\mathbb E}[M_t-M_s|{\cal F}_s]=0$.

{\it Case 2}. $m<k$. Then $t_m\le s<t_{m+1}\le t_k\le t<t_{k+1}$. So
\begin{eqnarray*}
I(t)-I(s)&=&\sum_{j=m}^{k-1}\Delta_{t_j}(M_{t_{j+1}}-M_{t_j})+\Delta_{t_k}(M_s-M_{t_k})-\Delta_{t_m}(M_s-M_{t_m})\\
&=&\sum_{j=m+1}^{k-1}\Delta_{t_j}(M_{t_{j+1}}-M_{t_j})+\Delta_{t_k}(M_t-M_{t_k})+\Delta_{t_m}(M_{t_{m+1}}-M_s).
\end{eqnarray*}
Hence
\begin{eqnarray*}
& &{\mathbb E}[I(t)-I(s)|{\cal
F}_s]\\
&=&\sum_{j=m+1}^{k-1}{\mathbb E}[\Delta_{t_j}{\mathbb E}[M_{t_{j+1}}-M_{t_j}|{\cal
F}_{t_j}]|{\cal F}_s]+{\mathbb E}[\Delta_{t_k}{\mathbb E}[M_t-M_{t_k}|{\cal
F}_{t_k}]|{\cal F}_s]+\Delta_{t_m}{\mathbb E}[M_{t_{m+1}}-M_s|{\cal F}_s]\\
&=&0.
\end{eqnarray*}

Combined, we can conclude $I(t)$ is a martingale.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.2.} Let $W(t)$, $0 \le t \le T$, be a Brownian motion, and let ${\cal F}(t)$, $0\le t \le T$, be an associated filtration. Let $\Delta(t)$, $0\le t \le T$, be a {\it nonrandom} simple process (i.e., there is a partition $\Pi = \{t_0, t_1, \cdots, t_n\}$ of $[0, T]$ such that for every $j$, $\Delta(t_j)$ is a nonrandom quantity and $\Delta(t) = \Delta(t_j)$ is constant in $t$ on the subinterval $[t_j, t_{j+1}))$. For $t \in [t_k, t_{k+1})$, define the stochastic integral
\[
I(t) = \sum_{j=0}^{k-1} \Delta(t_j) [W(t_{j+1}) - W(t_j)] + \Delta(t_k) [W(t) - W(t_k)].
\]

\smallskip

\noindent (i) Show that whenever $0 \le s < t \le T$, the increment $I(t) - I(s)$ is independent of ${\cal F}(s)$. (Simplification: If $s$ is between two partition points, we can always insert $s$ as an extra partition point. Then we can relabel the partition points so that they are still called $t_0$, $t_1$, $\cdots$, $t_n$, but with a larger value of $n$ and now with $s=t_k$ for some value of $k$. Of course, we must set $\Delta(s)=\Delta(t_{k-1})$ so that $\Delta$ takes the same value on the interval $[s, t_{k+1})$ as on the interval $[t_{k-1}, s)$. Similarly, we can insert $t$ as an extra partition point if it is not already one. Consequently, to show that $I(t)-I(s)$ is independent of ${\cal F}(s)$ for all $0 \le s < t \le T$, it suffices to show that $I(t_k)-I(t_l)$ is independent of ${\cal F}_l$ whenever $t_k$ and $t_l$ are two partition points with $t_l < t_k$. This is all you need to do.)
\begin{proof}
We follow the simplification in the hint and consider
$I(t_k)-I(t_l)$ with $t_l<t_k$. Then
$I(t_k)-I(t_l)=\sum_{j=l}^{k-1}\Delta(t_j)[W(t_{j+1})-W({t_j})]$.
Since $\Delta(t)$ is a non-random process and
$W(t_{j+1})-W(t_j)\perp {\cal F}(t_j) \supset {\cal F}({t_l})$ for
$j\ge l$, we must have $I(t_k)-I(t_l)\perp {\cal F}(t_l)$.
\end{proof}

\noindent (ii) Show that whenever $0 \le s < t \le T$, the increment $I(t) - I(s)$ is a normally distributed random variable with mean zero and variance $\int_s^t \Delta^2(u) du$.

\begin{proof} We use the notation in (i) and it is clear
that $I(t_k)-I(t_l)$ is normal since it is a linear combination of
independent normal random variables. Furthermore,
${\mathbb E}[I(t_k)-I(t_l)]=\sum_{j=l}^{k-1}\Delta_{t_j}{\mathbb E}[W(t_{j+1})-W(t_j)]=0$
and
$\mbox{Var}(I(t_k)-I(t_l))=\sum_{j=l}^{k-1}\Delta^2(t_j)\mbox{Var}[W(t_{j+1})-W(t_j)]=\sum_{j=l}^{k-1}\Delta^2(t_j)(t_{j+1}-t_j)=\int_{t_l}^{t_k}\Delta_u^2du
$.
\end{proof}

\noindent (iii) Use (i) and (ii) to show that $I(t)$, $0 \le t \le T$, is a martingale.

\begin{proof} By (i), ${\mathbb E}[I(t)-I(s)|{\cal F}(s)]={\mathbb E}[I(t)-I(s)]=0$, for
$s<t$, where the last equality is due to (ii).
\end{proof}

\noindent (iv) Show that $I^2(t) - \int_0^t \Delta^2(u) du$, $0\le t \le T$, is a martingale.

\begin{proof} For $s<t$,
\begin{eqnarray*}
& &{\mathbb E} \left[\left.I^2(t)-\int_0^t\Delta_u^2du-\left(I^2(s)-\int_0^s\Delta_u^2du\right)\right|{\cal
F}_s \right]\\
&=&{\mathbb E}\left[(I(t)-I(s))^2+2I(t)I(s)-2I^2(s)|{\cal
F}_s\right]-\int_s^t\Delta_u^2du\\
&=&{\mathbb E}[(I(t)-I(s))^2]+2I(s){\mathbb E}[I(t)-I(s)|{\cal
F}_s]-\int_s^t\Delta_u^2du\\
&=&\int_s^t\Delta_u^2du+0-\int_s^t\Delta_u^2du\\
&=&0.
\end{eqnarray*}
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.3.} We now consider a case in which $\Delta(t)$ in Exercise 4.2 is simple but random. In particular, let $t_0=0$, $t_1=s$, and $t_2=t$, and let $\Delta(0)$ be nonrandom and $\Delta(s)=W(s)$. Which of the following assertions is true?

\smallskip

\noindent (i)
$I(t)-I(s)$ is independent of ${\cal F}(s)$.
\begin{solution}We first note
\begin{eqnarray*}
I(t)-I(s) &=& \Delta(0)[W(t_1)-W(0)]+\Delta(t_1)[W(t_2)-W(t_1)]-\Delta(0)[W(t_1)-W(0)] \\
&=& \Delta(t_1)[W(t_2)-W(t_1)] \\
&=& W(s)[W(t)-W(s)].
\end{eqnarray*}
So $I(t)-I(s)$ is not independent of ${\cal F}(s)$, since $W(s) \in
{\cal F}(s)$.
\end{solution}

\noindent (ii) $I(t)-I(s)$ is normally distributed. (Hint: Check if the fourth moment is three times the square of the variance; see Exercise 3.3 of Chapter 3.)
\begin{solution} Following the hint, recall from (i), we know $I(t)-I(s) = W(s)[W(t)-W(s)]$. So
\[
{\mathbb E}[(I(t)-I(s))^4]={\mathbb E}[W_s^4]{\mathbb E}[(W_t-W_s)^4]=3s^2 \cdot
3(t-s)^2=9s^2(t-s)^2
\] and
\[
3 \left({\mathbb E}[(I(t)-I(s))^2]\right)^2=3 \left({\mathbb E}[W_s^2]{\mathbb E}[(W_t-W_s)^2]\right)^2=3s^2(t-s)^2.
\]
Since
${\mathbb E}[(I(t)-I(s))^4]\ne 3 \left({\mathbb E}[(I(t)-I(s))^2]\right)^2$, $I(t)-I(s)$ is not normally
distributed.
\end{solution}

\noindent (iii) ${\mathbb E}[I(t)|{\cal F}(s)] = I(s)$.
\begin{solution}
${\mathbb E}[I(t)-I(s)|{\cal F}(s)]=W(s){\mathbb E}[W(t)-W(s)|{\cal F}(s)]=0$. So ${\mathbb E}[I(t)|{\cal F}(s)] = I(s)$ is true.
\end{solution}

\noindent (iv) ${\mathbb E}[I^2(t) - \int_0^t \Delta^2(u)du | {\cal F}(s)] = I^2(s) - \int_0^s \Delta^2(u) du$.
\begin{solution}
\begin{eqnarray*} &
&{\mathbb E}\left[\left.I^2(t)-\int_0^t\Delta^2(u) du-\left(I^2(s)-\int_0^s\Delta^2(u)du\right)\right|{\cal
F}(s)\right]\\
&=&{\mathbb E}\left[(I(t)-I(s))^2+2I(t)I(s)-2I^2(s)-\left.\int_s^t\Delta^2(u)du\right|{\cal
F}(s)\right]\\
&=&{\mathbb E}\left[(I(t)-I(s))^2+2I(s)(I(t)-I(s))-\left.\int_s^tW^2(s)1_{(s,t]}(u)du\right|{\cal
F}(s)\right]\\
&=&{\mathbb E}[W^2(s)(W(t)-W(s))^2+2\Delta(0)W^2(s)(W(t)-W(s))-W^2(s)(t-s)|{\cal
F}(s)]\\
&=&W^2(s){\mathbb E}[(W(t)-W(s))^2]+2\Delta(0)W^2(s){\mathbb E}[W(t)-W(s)|{\cal F}(s)]-W^2(s)(t-s)\\
&=&W^2(s)(t-s)-W^2(s)(t-s)\\
&=&0.
\end{eqnarray*}
So ${\mathbb E}[I^2(t) - \int_0^t \Delta^2(u)du | {\cal F}(s)] = I^2(s) - \int_0^s \Delta^2(u) du$ is true.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.4 (Stratonovich integral).} Let $W(t)$, $t\ge 0$, be a Brownian motion. Let $T$ be a fixed positive number and let $\Pi=\{t_0, t_1, \cdots, t_n\}$ be a partition of $[0, T]$ (i.e., $0 = t_0 < t_1 < \cdots < t_n = T$).  For each $j$, define $t_j^*=\frac{t_j+t_{j+1}}{2}$ to be the midpoint of the interval $[t_j, t_{j+1}]$.

\smallskip

\noindent (i) Define the {\it half-sample quadratic variation} corresponding to $\Pi$ to be
\[
Q_{\Pi/2} = \sum_{j=0}^{n-1} (W(t_j^*)-W(t_j))^2.
\]
Show that $Q_{\Pi/2}$ has limit $\frac{1}{2}T$ as $|\!|\Pi|\!| \to 0$. (Hint: It suffices to show  ${\mathbb E} Q_{\Pi/2} = \frac{1}{2}T$ and $\lim_{|\!|\Pi|\!|\to 0} \mbox{Var}(Q_{\Pi/2})=0$.

\begin{proof} Following the hint, we first note that
\[
{\mathbb E}[Q_{\Pi/2}] = \sum_{j=0}^{n-1} {\mathbb E}[(W(t_j^*)-W(t_j))^2] = \sum_{j=0}^{n-1} (t_j^* - t_j) = \sum_{j=0}^{n-1} \frac{t_{j+1}-t_j}{2} = \frac{T}{2}.
\]
Then, by noting $W(t_j^*)-W(t_j)$ is equal to $W(t_j^*-t_j)=W\left(\frac{t_{j+1}-t_j}{2}\right)$ in distribution and ${\mathbb E}[(W^2(t)-t)^2]={\mathbb E}[W^4(t)-2tW^2(t)+t^2]=3{\mathbb E}[W^2(t)]^2-2t^2+t^2=2t^2$, we have
\begin{eqnarray*}
& & \mbox{Var}(Q_{\Pi/2})\\
&=&{\mathbb E}\left[\left(\sum_{j=0}^{n-1}\left(W(t_j^*)-W(t_j)\right)^2-\frac{T}{2}\right)^2\right]\\
&=&{\mathbb E}\left[\left(\sum_{j=0}^{n-1}\left(W(t_j^*)-W(t_j)\right)^2-\sum_{j=0}^{n-1}\frac{t_{j+1}-t_j}{2}\right)^2\right]\\
&=&\sum_{j,\,k=0}^{n-1}{\mathbb E}\left[\left(\left(W(t_j^*)-W(t_j)\right)^2-\frac{t_{j+1}-t_j}{2}\right)\left(\left(W(t_k^*)-W(t_k)\right)^2-\frac{t_{k+1}-t_k}{2}\right)\right]\\
&=& \sum_{j,\,k=0,j\ne k}^{n-1}{\mathbb E}\left[\left(\left(W(t_j^*)-W(t_j)\right)^2-\frac{t_{j+1}-t_j}{2}\right)\left(\left(W(t_k^*)-W(t_k)\right)^2-\frac{t_{k+1}-t_k}{2}\right)\right]\\
& & + \sum_{j=0}^{n-1}{\mathbb E}\left[\left((W(t_j^*)-W(t_j))^2-\frac{t_{j+1}-t_j}{2}\right)^2\right]\\
&=& 0 + \sum_{j=0}^{n-1}{\mathbb E}\left[\left(W^2(t_j^*-t_j)-\frac{t_{j+1}-t_j}{2}\right)^2\right] \\
&=&\sum_{j=0}^{n-1} 2\cdot\left(\frac{t_{j+1}-t_j}{2}\right)^2\\
&\le&\frac{T}{2}\max_{1\le j\le n}|t_{j+1}-t_j|\to 0.
\end{eqnarray*}
Combined, we can conclude $\lim_{|\!|\Pi|\!|\to 0} Q_{\Pi/2} \to \frac{T}{2}$ in ${\cal L}^2({\mathbb P})$.
\end{proof}

\noindent (ii) Define the Stratonovich integral of $W(t)$ with respect to $W(t)$ to be
\[
\int_0^T W(t) \circ dW(t) = \lim_{|\!|\Pi|\!|\to 0} \sum_{j=0}^{n-1} W(t_j^*) (W(t_{j+1})-W(t_j)). \tag{4.10.1}
\]
In contrast to the It\^{o} integral $\int_0^T W(t) dW(t) = \frac{1}{2}W^2(T) - \frac{1}{2}T$ of (4.3.4), which evaluates the integrand at the left endpoint of each subinterval $[t_j, t_{j+1}]$, here we evaluate the integrand at the midpoint $t_j^*$. Show that
\[
\int_0^T W(t) \circ dW(t) = \frac{1}{2} W^2(T).
\]
(Hint: Write the approximating sum in (4.10.1) as the sum of an approximating sum for the It\^{o} integral $\int_0^T W(t) dW(t)$ and $Q_{\Pi/2}$. The approximating sum for the It\^{o} integral is the one corresponding to the partition $0=t_0<t_0^*<t_1<t_1^*<\cdots<t_{n-1}^*<t_n=T$, not the partition $\Pi$.)

\begin{proof}
Following the hint and using the result in (i), we have
\begin{eqnarray*}
& & \sum_{j=0}^{n-1} W(t_j^*) (W(t_{j+1}) - W(t_j)) \\
&=& \sum_{j=0}^{n-1} W(t_j^*) \left\{[W(t_{j+1})-W(t_j^*)] + [W(t_j^*) - W(t_j)]\right\} \\
&=& \sum_{j=0}^{n-1} \left\{W(t_j^*)[W(t_{j+1})-W(t_j^*)] + W(t_j)[W(t_j^*)-W(t_j)] \right\} + \sum_{j=0}^{n-1} [W(t_j^*)-W(t_j)]^2.
\end{eqnarray*}
By the construction of It\^{o} integral,
\[
\lim_{|\!|\Pi^*|\!| \to 0}\sum_{j=0}^{n-1} \left\{W(t_j^*)[W(t_{j+1})-W(t_j^*)] + W(t_j)[W(t_j^*)-W(t_j)] \right\}  = \int_0^TW(t)dW(t),
\]
where $\Pi^*$ is the partition $0=t_0<t_0^*<t_1<t_1^*<\cdots<t_{n-1}^*<t_n=T$. (i) already shows
\[
\lim_{|\!|\Pi|\!|}\sum_{j=0}^{n-1} [W(t_j^*)-W(t_j)]^2=\frac{T}{2}=\frac{T}{2}.
\]
Combined, we conclude
\[
\int_0^T W(t) \circ dW(t) = \lim_{|\!|\Pi|\!|\to 0} \sum_{j=0}^{n-1} W(t_j^*) (W(t_{j+1})-W(t_j)) = \int_0^TW(t)dW(t) + \frac{T}{2}=\frac{1}{2}W^2(T).
\]
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.5 (Solving the generalized geometric Brownian motion equation).} Let $S(t)$ be a positive stochastic process that satisfies the generalized geometric Brownian motion differential equation (see Example 4.4.8)
\[
dS(t) = \alpha(t) S(t) dt + \sigma(t) S(t) dW(t), \tag{4.10.2}
\]
where $\alpha(t)$ and $\sigma(t)$ are processes adapted to the filtration ${\cal F}(t)$, $t\ge 0$, associated with the Brownian motion $W(t)$, $t\ge 0$. In this exercise, we show that $S(t)$ must be given by formula (4.4.26) (i.e., that formula provides the only solution to the stochastic differential equation (4.10.2)). In the process, we provide a method for solving this equation.

\smallskip

\noindent (i) Using (4.10.2) and the It\^{o}-Doeblin formula, compute $d\log S(t)$. Simplify so that you have a formula for $d\log S(t)$ that does not involve $S(t)$.
\begin{solution}
\[
d\log S_t=\frac{dS_t}{S_t}-\frac{1}{2}\frac{d\langle S
\rangle_t}{S_t^2}=\frac{2S_tdS_t-d\langle
S\rangle_t}{2S_t^2}=\frac{2S_t(\alpha_tS_tdt+\sigma_tS_tdW_t)-\sigma_t^2S_t^2dt}{2S_t^2}=\sigma_tdW_t+ \left(\alpha_t-\frac{1}{2}\sigma_t^2\right)dt.
\]
\end{solution}

\noindent (ii) Integrate the formula you obtained in (i), and then exponentiate the answer to obtain (4.4.26).
\begin{solution}
\[
\log S_t=\log
S_0+\int_0^t\sigma_sdW_s+\int_0^t \left(\alpha_s-\frac{1}{2}\sigma_s^2\right) ds.
\]
So $S_t=S_0\exp \left\{\int_0^t\sigma_sdW_s+\int_0^t \left(\alpha_s-\frac{1}{2}\sigma_s^2\right) ds\right\}$.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.6.} Let $S(t) = S(0) \exp \left\{\sigma W(t) + \left(\alpha-\frac{1}{2}\sigma^2\right)t\right\}$ be a geometric Brownian motion. Let $p$ be a positive constant. Compute $d(S^p(t))$, the differential of $S(t)$ raised to the power $p$.

\begin{solution} Without loss of generality, we assume
$p\ne 1$. Since $(x^p)'=px^{p-1}$, $(x^p)''=p(p-1)x^{p-2}$, we have
\begin{eqnarray*}
d(S_t^p)&=&pS_t^{p-1}dS_t+\frac{1}{2}p(p-1)S_t^{p-2}d\langle S
\rangle_t\\
&=&pS_t^{p-1}(\alpha S_tdt+\sigma
S_tdW_t)+\frac{1}{2}p(p-1)S_t^{p-2}\sigma^2S_t^2dt\\
&=&S_t^p \left[p\alpha dt+p\sigma dW_t+\frac{1}{2}p(p-1)\sigma^2dt \right]\\
&=&p S_t^p \left[\sigma dW_t+ \left(\alpha+\frac{p-1}{2}\sigma^2\right)dt \right].
\end{eqnarray*}
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.7.}

\noindent (i) Compute $dW^4(t)$ and then write $W^4(T)$ as the sum of an ordinary (Lebesgue) integral with respect to time and an It\^{o} integral.

\begin{solution} $dW_t^4=4W_t^3dW_t+\frac{1}{2}\cdot
4\cdot 3W_t^2d\langle W \rangle_t=4W_t^3dW_t+6W_t^2dt$. So
$W_T^4=4\int_0^TW_t^3dW_t+6\int_0^TW_t^2dt$.
\end{solution}

\noindent (ii) Take expectations on both sides of the formula you obtained in (i), use the fact that ${\mathbb E}W^2(t) = t$, and derive the formula ${\mathbb E}W^4(T)=3T^2$.

\begin{solution} ${\mathbb E}[W_T^4]=4{\mathbb E}\left[\int_0^TW_t^3dW_t\right]+6\int_0^T{\mathbb E}[W_t^2]dt=6\int_0^Ttdt=3T^2$.
\end{solution}

\noindent (iii) Use the method of (i) and (ii) to derive a formula for ${\mathbb E}W^6(T)$.

\begin{solution}$dW_t^6=6W_t^5dW_t+\frac{1}{2}\cdot 6\cdot 5
W_t^4dt$. So $W_T^6=6\int_0^TW_t^5dW_t+15\int_0^TW_t^4dt$. Hence
${\mathbb E}[W_T^6]=15\int_0^T3t^2dt=15T^3$.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.8 (Solving the Vasicek equation).} The Vasicek interest rate stochastic differential equation (4.4.32) is
\[
dR(t) = (\alpha - \beta R(t))dt + \sigma dW(t),
\]
where $\alpha$, $\beta$, and $\sigma$ are positive constants. The solution to this equation is given in Example 4.4.10. This exercise shows how to derive this solution.

\smallskip

\noindent (i) Use (4.4.32) and the It\^{o}-Doeblin formula to compute $d(e^{\beta t} R(t))$. Simplify it so that you have a formula for $d(e^{\beta t}R(t))$ that does not involve $R(t)$.

\begin{solution}
$d(e^{\beta t}R_t)=\beta e^{\beta
t}R_tdt+e^{\beta t}dR_t=e^{\beta t}[\beta R_t dt + (\alpha - \beta R(t))dt + \sigma dW(t)] = e^{\beta}(\alpha dt + \sigma dW(t))$.
\end{solution}

\noindent (ii) Integrate the equation you obtain in (i) and solve for $R(t)$ to obtain (4.4.33).

\begin{solution}
\[
e^{\beta t}R_t=R_0+\int_0^te^{\beta s}(\alpha ds+\sigma
dW_s)=R_0+\frac{\alpha}{\beta}(e^{\beta t}-1)+\sigma\int_0^te^{\beta
s}dW_s.\]
Therefore $R_t=R_0e^{-\beta t}+\frac{\alpha}{\beta}(1-e^{-\beta
t})+\sigma\int_0^te^{-\beta(t-s)}dW_s$.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.9.} For a European call expiring at time $T$ with strike price $K$, the Black-Scholes-Merton price at time $t$, if the time-$t$ stock price is $x$, is
\[
c(t,x) = xN(d_+(T-t,x)) - Ke^{-r(T-t)}N(d_-(T-t,x)),
\]
where
\begin{eqnarray*}
d_+(\tau,x) &=& \frac{1}{\sigma\sqrt{\tau}} \left[\log \frac{x}{K} + \left(r+\frac{1}{2}\sigma^2\right) \tau \right] \\
d_-(\tau,x) &=& d_+(\tau,x) - \sigma\sqrt{\tau},
\end{eqnarray*}
and $N(y)$ is the cumulative standard normal distribution
\[
N(y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^y e^{-\frac{z^2}{2}}dz = \frac{1}{\sqrt{2\pi}} \int_{-y}^{\infty} e^{-\frac{z^2}{2}}dz.
\]
The purpose of this exercise is to show that the function $c$ satisfies the Black-Scholes-Merton partial differential equation
\[
c_t(t,x) + rxc_x(t,x) + \frac{1}{2}\sigma^2x^2c_{xx}(t,x) = rc(t,x), 0 \le t < T, x>0,
\tag{4.10.3}
\]
the {\it terminal condition}
\[
\lim_{t\uparrow T} c(t,x) = (x-K)^+, x>0, x\ne K, \tag{4.10.4}
\]
and the {\it boundary conditions}
\[
\lim_{x\downarrow 0} c(t,x) = 0, \lim_{x\to\infty} [c(t,x)-(x-e^{-r(T-t)}K)]=0, 0 \le t < T.
\tag{4.10.5}
\]
Equation (4.10.4) and the first part of (4.10.5) are usually written more simply but less precisely as
\[
c(T,x) = (x-K)^+, x\ge 0
\]
and
\[
c(t,0) = 0, 0 \le t \le T.
\]

For this exercise, we abbreviate $c(t,x)$ as simply $c$ and $d_{\pm}(T-t,x)$ as simply $d_{\pm}$.

\smallskip

\noindent (i) Verify first the equation
\[
Ke^{-r(T-t)}N'(d_-) = x N'(d_+). \tag{4.10.6}
\]

\begin{proof}
\begin{eqnarray*}
Ke^{-r(T-t)}N'(d_-) &=&
Ke^{-r(T-t)}\frac{e^{-\frac{d^2_-}{2}}}{\sqrt{2\pi}}\\
&=&Ke^{-r(T-t)}\frac{e^{-\frac{(d_+-\sigma\sqrt{T-t})^2}{2}}}{\sqrt{2\pi}}\\
&=&Ke^{-r(T-t)}e^{\sigma\sqrt{T-t}d_+}e^{-\frac{\sigma^2(T-t)}{2}}N'(d_+)\\
&=&Ke^{-r(T-t)}\frac{x}{K}e^{(r+\frac{\sigma^2}{2})(T-t)}e^{-\frac{\sigma^2(T-t)}{2}}N'(d_+)\\
&=&x N'(d_+).
\end{eqnarray*}
\end{proof}

\noindent (ii) Show that $c_x = N(d_+)$. This is the {\it delta} of the option. (Be careful! Remember that $d_+$ is a function of $x$.)

\begin{proof} By equation (4.10.6) from (i),
\begin{eqnarray*}
c_x&=&N(d_+)+xN'(d_+)\frac{\partial}{\partial
x}d_+(T-t,x)-Ke^{-r(T-t)}N'(d_-)\frac{\partial}{\partial
x}d_-(T-t,x)\\
&=&N(d_+)+xN'(d_+)\frac{\partial}{\partial x}d_+'(T-t,x)-xN'(d_+)\frac{\partial}{\partial x}d_+(T-t,x)\\
&=&N(d_+).
\end{eqnarray*}
\end{proof}

\noindent (iii) Show that
\[
c_t = -rKe^{-r(T-t)}N(d_-)-\frac{\sigma x}{2\sqrt{T-t}}N'(d_+).
\]
This is the {\it theta} of the option.

\begin{proof}
\begin{eqnarray*}
c_t&=&xN'(d_+)\frac{\partial }{\partial
t}d_+(T-t,x)-rKe^{-r(T-t)}N(d_-)-Ke^{-r(T-t)}N'(d_-)\frac{\partial}{\partial
t}d_-(T-t,x)\\
&=&xN'(d_+)\frac{\partial}{\partial
t}d_+(T-t,x)-rKe^{-r(T-t)}N(d_-)-xN'(d_+)\left[\frac{\partial}{\partial
t}d_+(T-t,x)+\frac{\sigma}{2\sqrt{T-t}}\right]\\
&=&-rKe^{-r(T-t)}N(d_-)-\frac{\sigma x}{2\sqrt{T-t}}N'(d_+).
\end{eqnarray*}
\end{proof}

\noindent (iv) Use the formulas above to show that $c$ satisfies (4.10.3).

\begin{proof}
\begin{eqnarray*}
& &c_t+rxc_x+\frac{1}{2}\sigma^2x^2c_{xx}\\
&=&-rKe^{-r(T-t)}N(d_-)-\frac{\sigma
x}{2\sqrt{T-t}}N'(d_+)+rxN(d_+)+\frac{1}{2}\sigma^2x^2N'(d_+)\frac{\partial
}{\partial x}d_+(T-t,x)\\
&=&rc-\frac{\sigma
x}{2\sqrt{T-t}}N'(d_+)+\frac{1}{2}\sigma^2x^2N'(d_+)\frac{1}{\sigma\sqrt{T-t}x}\\
&=&rc.
\end{eqnarray*}
\end{proof}

\noindent (v) Show that for $x>K$, $\lim_{t\uparrow T}d_{\pm}=\infty$, but for $0<x<K$, $\lim_{t\uparrow T} d_{\pm} = -\infty$. Use these equalities to derive the terminal condition (4.10.4).

\begin{proof} For $x>K$, $d_+(T-t,x)>0$ and $\lim_{t\uparrow
T}d_+(T-t,x)=\lim_{\tau\downarrow 0}d_+(\tau,x)=\infty$.
$\lim_{t\uparrow T}d_-(T-t,x)=\lim_{\tau\downarrow
0}d_-(\tau,x)=\lim_{\tau\downarrow
0}\left(\frac{1}{\sigma\sqrt{\tau}}\ln\frac{x}{K}+\frac{1}{\sigma}(r+\frac{1}{2}\sigma^2)\sqrt{\tau}-\sigma\sqrt{\tau}\right)=\infty$.
Similarly, $\lim_{t\uparrow T}d_{\pm}=-\infty$ for $x\in (0,K)$.
Also it's clear that $\lim_{t\uparrow T}d_{\pm}=0$ for $x=K$. So
\[
\lim_{t\uparrow T}c(t,x)=xN(\lim_{t\uparrow
T}d_+)-KN(\lim_{t\uparrow T}d_-)=\begin{cases} x-K,& \mbox{if $x>K$}\\
0,&\mbox{if $x\le K$}
\end{cases}=(x-K)^+.
\]
\end{proof}

\noindent (vi) Show that for $0\le t <T$, $\lim_{x\downarrow 0}d_{\pm} = -\infty$. Use this fact to verify the first part of boundary condition (4.10.5) as $x\downarrow 0$.

\begin{proof} We note
\[
\lim_{x\downarrow 0} d_+ = \frac{1}{\sigma\sqrt{\tau}} \left[\lim_{x\downarrow 0}\log\frac{x}{K} + \left(r+\frac{1}{2}\sigma^2\right)\tau\right] = -\infty
\]
and
\[
\lim_{x\downarrow 0} d_- = \lim_{x\downarrow 0} d_+ - \sigma\sqrt{\tau} = -\infty.
\]
So for $t\in [0,T)$,
\[
\lim_{x\downarrow
0}c(t,x)=\lim_{x\downarrow
0}xN(\lim_{x\downarrow 0}d_{+}(T-t,x))-Ke^{-r(T-t)}N(\lim_{x\downarrow
0}d_-(T-t,x))=0.
\]
\end{proof}

\noindent (vii) Show that for $0 \le t <T$, $\lim_{x\to\infty} d_{\pm} = \infty$. Use this fact to verify the second part of boundary condition (4.10.5) as $x\to\infty$. In this verification, you will need to show that
\[
\lim_{x\to\infty} \frac{N(d_+)-1}{x^{-1}} = 0.
\]
This is an indeterminate form $\frac{0}{0}$, and L'H\^{o}spital's rule implies that this limit is
\[
\lim_{x\to\infty}\frac{\frac{d}{dx} \left[N(d_+)-1\right]}{\frac{d}{dx}x^{-1}}.
\]
Work out this expression and use the fact that
\[
x = K \exp \left\{\sigma\sqrt{T-t} d_+ - (T-t) \left(r+\frac{1}{2}\sigma^2\right)\right\}
\]
to write this expression solely in terms of $d_+$ (i.e., without the appearance of any $x$ except the $x$ in the argument of $d_+(T-t,x)$). Then argue that the limit is zero as $d_+\to\infty$.

\begin{proof}For $t\in [0,T)$, it is clear
$\lim_{x\to\infty}d_{\pm}=\infty$. Following the hint, we note
\[
\lim_{x\to\infty}x(N(d_+)-1)=\lim_{x\to\infty}\frac{N'(d_+)\frac{\partial}{\partial
x}d_+}{-x^{-2}}=\lim_{x\to\infty}\frac{N'(d_+)\frac{1}{\sigma\sqrt{T-t}}}{-x^{-1}}.\]
By the expression of $d_+$, we have
$x=K\exp\left\{\sigma\sqrt{T-t}d_+-(T-t)(r+\frac{1}{2}\sigma^2)\right\}$. So
\[
\lim_{x\to\infty}x(N(d_+)-1)=\lim_{x\to\infty}N'(d_+)\frac{-x}{\sigma\sqrt{T-t}}=\lim_{d_+\to\infty}\frac{e^{-\frac{d^2_+}{2}}}{\sqrt{2\pi}}\frac{-Ke^{\sigma\sqrt{T-t}d_+-(T-t)(r+\frac{1}{2}\sigma^2)}}{\sigma\sqrt{T-t}}=0.
\]
Therefore
\begin{eqnarray*}
& &\lim_{x\to\infty}[c(t,x)-(x-e^{-r(T-t)}K)]\\
&=&\lim_{x\to\infty}[x N(d_+)-Ke^{-r(T-t)N(d_-)}-x+Ke^{-r(T-t)}]\\
&=&\lim_{x\to\infty}[x(N(d_+)-1)+Ke^{-r(T-t)}(1-N(d_-))]\\
&=&\lim_{x\to\infty}x(N(d_+)-1)+Ke^{-r(T-t)}(1-N(\lim_{x\to\infty}d_-))\\
&=&0.
\end{eqnarray*}
\end{proof}

\medskip

\noindent ****************************** UPDATE STOPPED HERE *************************************

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.10 (Self-financing trading).} The fundamental idea behind no-arbitrage pricing is to reproduce the payoff of a derivative security by trading in the underlying asset (which we call a stock) and the money market account. In discrete time, we let $X_k$ denote the value of the hedging portfolio at time $k$ and let $\Delta_k$ denote the number of shares of stock held between times $k$ and $k+1$. Then, at time $k$, after rebalancing (i.e., moving from a position of $\Delta_{k-1}$ to a position $\Delta_k$ in the stock), the amount in the money market account is $X_k - S_k \Delta_k$. The value of the portfolio at time $k+1$ is
\[
X_{k+1} = \Delta_k S_{k+1} + (1+r) (X_k - \Delta_k S_k). \tag{4.10.7}
\]
This formula can be rearranged to become
\[
X_{k+1} -  X_k = \Delta_k (S_{k+1} - S_k) + r (X_k - \Delta_k S_k), \tag{4.10.8}
\]
which says that the gain between time $k$ and time $k+1$ is the sum of the capital gain on the stock holdings, $\Delta_k(S_{k+1}-S_k)$, and the interest earnings on the money market account, $r(X_k - \Delta_k S_k)$. The continuous-time analogue of (4.10.8) is
\[
dX(t) = \Delta(t) dS(t) + r(X(t) - \Delta(t)S(t))dt. \tag{4.10.9}
\]

Alternatively, one could define the value of a share of the money market account at time $k$ to be
\[
M_k = (1+r)^k
\]
and formulate the discrete-time model with two processes, $\Delta_k$ as before and $\Gamma_k$ denoting the number of shares of the money market account held at time $k$ after rebalancing. Then
\[
X_k = \Delta_k S_k + \Gamma_k M_k, \tag{4.10.10}
\]
so that (4.10.7) becomes
\[
X_{k+1} = \Delta_k S_{k+1} + (1+r)\Gamma_k M_k = \Delta_k S_{k+1} + \Gamma_k M_{k+1}. \tag{4.10.11}
\]
Subtracting (4.10.10) from (4.10.11), we obtain in place of (4.10.8) the equation
\[
X_{k+1} - X_k = \Delta_k (S_{k+1}-S_k)+\Gamma_k(M_{k+1}-M_k), \tag{4.10.12}
\]
which says that the gain between time $k$ and time $k+1$ is the sum of the capital gain on stock holdings, $\Delta_k (S_{k+1}-S_k)$, and the earnings from the money market investment, $\Gamma_k(M_{k+1}-M_k)$.

But $\Delta_k$ and $\Gamma_k$ cannot be chosen arbitrarily. The agent arrives at time

\smallskip

\noindent (i)\begin{proof} We show $(4.10.16)+(4.10.9)
\Longleftrightarrow (4.10.16)+(4.10.15)$, i.e. assuming $X$ has the
representation $X_t=\Delta_tS_t+\Gamma_tM_t$, ``continuous-time
self-financing condition" has two equivalent formulations,
$(4.10.9)$ or $(4.10.15)$. Indeed,
$dX_t=\Delta_tdS_t+\Gamma_tdM_t+(S_td\Delta_t+dS_td\Delta_t+M_td\Gamma_t+dM_td\Gamma_t)$.
So $dX_t=\Delta_tdS_t+\Gamma_tdM_t \Longleftrightarrow
S_td\Delta_t+dS_td\Delta_t+M_td\Gamma_t+dM_td\Gamma_t=0$, i.e.
$(4.10.9)\Longleftrightarrow (4.10.15)$.
\end{proof}

(ii)\begin{proof} First, we clarify the problems by stating
explicitly the given conditions and the result to be proved. We
assume we have a portfolio $X_t=\Delta_tS_t+\Gamma_tM_t$. We let
$c(t,S_t)$ denote the price of call option at time $t$ and set
$\Delta_t=c_x(t,S_t)$. Finally, we assume the portfolio is
self-financing. The problem is to show
\[
rN_tdt=\left[c_t(t,S_t)+\frac{1}{2}\sigma^2S_t^2c_{xx}(t,S_t)\right]dt,
\]where $N_t=c(t,S_t)-\Delta_tS_t$.

Indeed, by the self-financing property and $\Delta_t=c_x(t,S_t)$, we
have $c(t,S_t)=X_t$ (by the calculations in Subsection 4.5.1-4.5.3).
This uniquely determines $\Gamma_t$ as
\[
\Gamma_t=\frac{X_t-\Delta_tS_t}{M_t}=\frac{c(t,S_t)-c_x(t,S_t)S_t}{M_t}=\frac{N_t}{M_t}.
\]Moreover,
\begin{eqnarray*}
dN_t&=&\left[c_t(t,S_t)dt+c_x(t,S_t)dS_t+\frac{1}{2}c_{xx}(t,S_t)d\langle S_t \rangle_t\right]-d(\Delta_tS_t)\\
&=&\left[c_t(t,S_t)+\frac{1}{2}c_{xx}(t,S_t)\sigma^2S_t^2\right]dt+[c_x(t,S_t)dS_t-d(X_t-\Gamma_tM_t)]\\
&=&\left[c_t(t,S_t)+\frac{1}{2}c_{xx}(t,S_t)\sigma^2S_t^2\right]dt+M_td\Gamma_t+dM_td\Gamma_t+[c_x(t,S_t)dS_t+\Gamma_tdM_t-dX_t].
\end{eqnarray*}
By self-financing property,
$c_x(t,S_t)dt+\Gamma_tdM_t=\Delta_tdS_t+\Gamma_tdM_t=dX_t$, so
\[
\left[c_t(t,S_t)+\frac{1}{2}c_{xx}(t,S_t)\sigma^2S_t^2\right]dt=dN_t-M_td\Gamma_t-dM_td\Gamma_t=\Gamma_tdM_t=\Gamma_trM_tdt=rN_tdt.
\]
\end{proof}

\medskip

\noindent 4.11.\begin{proof}First, we note $c(t,x)$ solves the
Black-Scholes-Merton PDE with volatility $\sigma_1$:
\[
\left(\frac{\partial}{\partial t}+rx\frac{\partial}{\partial
x}+\frac{1}{2}x^2\sigma_1^2\frac{\partial^2}{\partial
x^2}-r\right)c(t,x)=0.
\]
So
\[
c_t(t,S_t)+rS_tc_x(t,S_t)+\frac{1}{2}\sigma_1^2S_t^2c_{xx}(t,S_t)-rc(t,S_t)=0,
\]and
\begin{eqnarray*}
dc(t,S_t)&=&c_t(t,S_t)dt+c_x(t,S_t)(\alpha
S_tdt+\sigma_2S_tdW_t)+\frac{1}{2}c_{xx}(t,S_t)\sigma_2^2S_t^2dt\\
&=&\left[c_t(t,S_t)+\alpha c_x(t,S_t)S_t+\frac{1}{2}\sigma_2^2S_t^2c_{xx}(t,S_t)\right]dt+\sigma_2S_tc_x(t,S_t)dW_t\\
&=&\left[rc(t,S_t)+(\alpha-r)c_x(t,S_t)S_t+\frac{1}{2}S_t^2(\sigma_2^2-\sigma_1^2)c_{xx}(t,S_t)\right]dt+\sigma_2S_tc_{x}(t,S_t)dW_t.
\end{eqnarray*}
Therefore
\begin{eqnarray*}
dX_t
&=&\left[rc(t,S_t)+(\alpha-r)c_x(t,S_t)S_t+\frac{1}{2}S_t^2(\sigma_2^2-\sigma_1^2)\sigma_{xx}(t,S_t)+rX_t-rc(t,S_t)+rS_tc_x(t,S_t)\right.\\
& &\left.-\frac{1}{2}
(\sigma_2^2-\sigma_1^2)S_t^2c_{xx}(t,S_t)-c_x(t,S_t)\alpha
S_t\right]dt+[\sigma_2S_tc_x(t,S_t)-c_x(t,S_t)\sigma_2S_t]dW_t\\
&=&rX_tdt.
\end{eqnarray*}
This implies $X_t=X_0e^{rt}$. By $X_0$, we conclude $X_t=0$ for all
$t\in [0,T]$.

\end{proof}

\medskip

\noindent 4.12. (i)
\begin{proof}
By (4.5.29), $c(t,x)-p(t,x)=x-e^{-r(T-t)}K$. So
$p_x(t,x)=c_x(t,x)-1=N(d_+(T-t,x))-1$,
$p_{xx}(t,x)=c_{xx}(t,x)=\frac{1}{\sigma x\sqrt{T-t}}N'(d_+(T-t,x))$
and
\begin{eqnarray*}
p_t(t,x)&=&c_t(t,x)+re^{-r(T-t)}K\\
&=&-rKe^{-r(T-t)}N(d_-(T-t,x))-\frac{\sigma
x}{2\sqrt{T-t}}N'(d_+(T-t,x))+rKe^{-r(T-t)}\\
&=&rKe^{-r(T-t)}N(-d_-(T-t,x))-\frac{\sigma
x}{2\sqrt{T-t}}N'(d_+(T-t,x)).
\end{eqnarray*}
\end{proof}

(ii)\begin{proof} For an agent hedging a short position in the put,
since $\Delta_t=p_x(t,x)<0$, he should short the underlying stock
and put $p(t,S_t)-p_x(t,S_t)S_t (>0)$ cash in the money market
account.
\end{proof}

(iii)\begin{proof} By the put-call parity, it suffices to show
$f(t,x)=x-Ke^{-r(T-t)}$ satisfies the Black-Scholes-Merton partial
differential equation. Indeed,
\[
\left(\frac{\partial}{\partial
t}+\frac{1}{2}\sigma^2x^2\frac{\partial^2}{\partial
x^2}+rx\frac{\partial}{\partial
x}-r\right)f(t,x)=-rKe^{-r(T-t)}+\frac{1}{2}\sigma^2x^2\cdot
0+rx\cdot 1-r(x-Ke^{-r(T-t)})=0.
\]

Remark: The Black-Scholes-Merton PDE has many solutions. Proper
boundary conditions are the key to uniqueness. For more details, see
Wilmott \cite{Wilmott95}.
\end{proof}

\medskip

\noindent 4.13. \begin{proof} We suppose $(W_1,W_2)$ is a pair of
local martingale defined by SDE
\begin{equation}
\begin{cases}
dW_1(t)=dB_1(t)\\
dW_2(t)=\alpha(t)dB_1(t)+\beta(t)dB_2(t).
\end{cases}
\end{equation}
We want to find $\alpha(t)$ and $\beta(t)$ such that
\begin{equation}
\begin{cases}
(dW_2(t))^2=[\alpha^2(t)+\beta^2(t)+2\rho(t)\alpha(t)\beta(t)]dt=dt\\
dW_1(t)dW_2(t)=[\alpha(t)+\beta(t)\rho(t)]dt=0.
\end{cases}
\end{equation}
Solve the equation for $\alpha(t)$ and $\beta(t)$, we have
$\beta(t)=\frac{1}{\sqrt{1-\rho^2(t)}}$ and
$\alpha(t)=-\frac{\rho(t)}{\sqrt{1-\rho^2(t)}}$. So
\begin{equation}
\begin{cases}
W_1(t)=B_1(t)\\
W_2(t)=\int_0^t\frac{-\rho(s)}{\sqrt{1-\rho^2(s)}}dB_1(s)+\int_0^t\frac{1}{\sqrt{1-\rho^2(s)}}dB_2(s)
\end{cases}
\end{equation}
is a pair of independent BM's. Equivalently, we have
\begin{equation}
\begin{cases}
B_1(t)=W_1(t)\\
B_2(t)=\int_0^t\rho(s)dW_1(s)+\int_0^t\sqrt{1-\rho^2(s)}dW_2(s).
\end{cases}
\end{equation}
\end{proof}

\medskip

\noindent 4.14. (i)
\begin{proof}
Clearly $Z_j\in {\cal F}_{t_{j+1}}$. Moreover
\[
E[Z_j|{\cal
F}_{t_j}]=f''(W_{t_j})E[(W_{t_{j+1}}-W_{t_j})^2-(t_{j+1}-t_j)|{\cal
F}_{t_j}]=f''(W_{t_j})(E[W^2_{t_{j+1}-t_j}]-(t_{j+1}-t_j))=0,
\]since $W_{t_{j+1}}-W_{t_j}$ is independent of ${\cal F}_{t_j}$ and
$W_t\thicksim N(0,t)$. Finally, we have
\begin{eqnarray*}
E[Z_j^2|{\cal F}_{t_j}] &=&
[f''(W_{t_j})]^2E[(W_{t_{j+1}}-W_{t_j})^4-2(t_{j+1}-t_j)(W_{t_{j+1}}-W_{t_j})^2+(t_{j+1}-t_j)^2|{\cal
F}_{t_j}]\\
&=&
[f''(W_{t_j})]^2(E[W^4_{t_{j+1}-t_j}]-2(t_{j+1}-t_j)E[W^2_{t_{j+1}-t_j}]+(t_{j+1}-t_j)^2)\\
&=&[f''(W_{t_j})]^2[3(t_{j+1}-t_j)^2-2(t_{j+1}-t_j)^2+(t_{j+1}-t_j)^2]\\
&=&2[f''(W_{t_j})]^2(t_{j+1}-t_j)^2,
\end{eqnarray*}
where we used the independence of Browian motion increment and the
fact that $E[X^4]=3E[X^2]^2$ if $X$ is Gaussian with mean 0.
\end{proof}

(ii)\begin{proof}$E[\sum_{j=0}^{n-1}Z_j]=E[\sum_{j=0}^{n-1}E[Z_j|{\cal
F}_{t_j}]]=0$ by part (i).\end{proof}

(iii)\begin{proof}
\begin{eqnarray*}
Var[\sum_{j=0}^{n-1}Z_j]
&=&E[(\sum_{j=0}^{n-1}Z_j)^2]\\
&=&E[\sum_{j=0}^{n-1}Z_j^2+2\sum_{0\le i<j\le
n-1}Z_iZ_j]\\
&=&\sum_{j=0}^{n-1}E[E[Z_j^2|{\cal F}_{t_j}]]+2\sum_{0\le i<j\le
n-1}E[Z_iE[Z_j|{\cal
F}_{t_j}]]\\
&=&\sum_{j=0}^{n-1}E[2[f''(W_{t_j})]^2(t_{j+1}-t_j)^2]\\
&=&\sum_{j=0}^{n-1}2E[(f''(W_{t_j}))^2](t_{j+1}-t_j)^2\\
&\le& 2\max_{0\le j\le n-1}|t_{j+1}-t_j|\cdot
\sum_{j=0}^{n-1}E[(f''(W_{t_j}))^2](t_{j+1}-t_j)\\
&\to& 0,
\end{eqnarray*}since
$\sum_{j=0}^{n-1}E[(f''(W_{t_j}))^2](t_{j+1}-t_j)\to\int_0^TE[(f''(W_t))^2]dt<\infty$.
\end{proof}

\medskip

\noindent 4.15. (i) \begin{proof} $B_i$ is a local martingale with
\[
(dB_i(t))^2=\left(\sum_{j=1}^d\frac{\sigma_{ij}(t)}{\sigma_i(t)}dW_j(t)\right)^2=\sum_{j=1}^d\frac{\sigma^2_{ij}(t)}{\sigma_i^2(t)}dt=dt.
\]So $B_i$ is a Brownian motion.
\end{proof}

(ii) \begin{proof}
\begin{eqnarray*}
dB_i(t)dB_k(t) &=&\left[\sum_{j=1}^d\frac{\sigma_{ij}(t)}{\sigma_i(t)}dW_j(t)\right]\left[\sum_{l=1}^d\frac{\sigma_{kl}(t)}{\sigma_k(t)}dW_l(t)\right]\\
&=&\sum_{1\le j,\;l\le d}\frac{\sigma_{ij}(t)\sigma_{kl}(t)}{\sigma_i(t)\sigma_k(t)}dW_j(t)dW_l(t)\\
&=&\sum_{j=1}^d\frac{\sigma_{ij}(t)\sigma_{kj}(t)}{\sigma_i(t)\sigma_k(t)}dt\\
&=&\rho_{ik}(t)dt.
\end{eqnarray*}
\end{proof}

\medskip

\noindent 4.16.\begin{proof}To find the $m$ independent Brownion
motion $W_1(t)$, $\cdots$, $W_m(t)$, we need to find
$A(t)=(a_{ij}(t))$ so that
\[
(dB_1(t),\cdots,dB_m(t))^{tr}=A(t)(dW_1(t),\cdots,dW_m(t))^{tr},
\]or equivalently
\begin{eqnarray*}
(dW_1(t),\cdots,dW_m(t))^{tr}=A(t)^{-1}(dB_1(t),\cdots,dB_m(t))^{tr},
\end{eqnarray*}
and
\begin{eqnarray*}
&
&(dW_1(t),\cdots,dW_m(t))^{tr}(dW_1(t),\cdots,dW_m(t))\\
&=&A(t)^{-1}(dB_1(t),\cdots,dB_m(t))^{tr}(dB_1(t),\cdots,dB_m(t))(A(t)^{-1})^{tr}dt\\
&=&I_{m\times m}dt,
\end{eqnarray*}
 where $I_{m\times m}$ is the $m\times m$ unit
matrix. By the condition $dB_i(t)dB_k(t)=\rho_{ik}(t)dt$, we get
\[
(dB_1(t),\cdots,dB_m(t))^{tr}(dB_1(t),\cdots,dB_m(t))=C(t).
\]So $A(t)^{-1}C(t)(A(t)^{-1})^{tr}=I_{m\times m}$, which gives
$C(t)=A(t)A(t)^{tr}$. This motivates us to define $A$ as the square
root of $C$. Reverse the above analysis, we obtain a formal proof.
\end{proof}

\medskip

\noindent 4.17.
\begin{proof}
We will try to solve all the sub-problems in a single, long
solution. We start with the general $X_i$:
\[
X_i(t)=X_i(0)+\int_0^t\theta_i(u)du+\int_0^t\sigma_i(u)dB_i(u),\;i=1,2.
\]The goal is to show
\[
\lim_{\epsilon\downarrow
0}\frac{C(\epsilon)}{\sqrt{V_1(\epsilon)V_2(\epsilon)}}=\rho(t_0).
\]

First, for $i=1,2$, we have
\begin{eqnarray*}
M_i(\epsilon)&=&E[X_i(t_0+\epsilon)-X_i(t_0)|{\cal
F}_{t_0}]\\
&=&E\left[\int_{t_0}^{t_0+\epsilon}\Theta_i(u)du+\int_{t_0}^{t_0+\epsilon}\sigma_i(u)dB_i(u)|{\cal
F}_{t_0}\right]\\
&=&\Theta_i(t_0)\epsilon+E\left[\int_{t_0}^{t_0+\epsilon}(\Theta_i(u)-\Theta_i(t_0))du|{\cal
F}_{t_0}\right]. \end{eqnarray*} By Conditional Jensen's Inequality,
\[
\left|
E\left[\int_{t_0}^{t_0+\epsilon}(\Theta_i(u)-\Theta_i(t_0))du|{\cal
F}_{t_0}\right]\right| \le
E\left[\int_{t_0}^{t_0+\epsilon}|\Theta_i(u)-\Theta_i(t_0)|du|{\cal
F}_{t_0}\right]
\]
Since
$\frac{1}{\epsilon}\int_{t_0}^{t_0+\epsilon}|\Theta_i(u)-\Theta_i(t_0)|du
\le 2M$ and $\lim_{\epsilon\to
0}\frac{1}{\epsilon}\int_{t_0}^{t_0+\epsilon}|\Theta_i(u)-\Theta_i(t_0)|du=0$
by the continuity of $\Theta_i$, the Dominated Convergence Theorem
under Conditional Expectation implies
\[
\lim_{\epsilon\to
0}\frac{1}{\epsilon}E\left[\int_{t_0}^{t_0+\epsilon}|\Theta_i(u)-\Theta_i(t_0)|du|{\cal
F}_{t_0}\right]=E\left[\lim_{\epsilon\to
0}\frac{1}{\epsilon}\int_{t_0}^{t_0+\epsilon}|\Theta_i(u)-\Theta_i(t_0)|du|{\cal
F}_{t_0}\right]=0.
\]
So $M_i(\epsilon)=\Theta_i(t_0)\epsilon+o(\epsilon)$. This proves
(iii).

To calculate the variance and covariance, we note
$Y_i(t)=\int_0^t\sigma_i(u)dB_i(u)$ is a martingale and by It\^{o}'s
formula $Y_i(t)Y_j(t)-\int_0^t\sigma_i(u)\sigma_j(u)du$ is a
martingale $(i=1,2)$. So
\begin{eqnarray*}
& &E[(X_i(t_0+\epsilon)-X_i(t_0))(X_j(t_0+\epsilon)-X_j(t_0))|{\cal F}_{t_0}]\\
&=&E\left[\left(Y_i(t_0+\epsilon)-Y_i(t_0)+\int_{t_0}^{t_0+\epsilon}\Theta_i(u)du\right)\left(Y_j(t_0+\epsilon)-Y_j(t_0)+\int_{t_0}^{t_0+\epsilon}\Theta_j(u)du\right)|{\cal
F}_{t_0}\right]\\
&=&E\left[(Y_i(t_0+\epsilon)-Y_i(t_0))\left(Y_j(t_0+\epsilon)-Y_j(t_0)\right)|{\cal
F}_{t_0}\right]+E\left[\int_{t_0}^{t_0+\epsilon}\Theta_i(u)du\int_{t_0}^{t_0+\epsilon}\Theta_j(u)du|{\cal
F}_{t_0}\right]\\
&
&+E\left[(Y_i(t_0+\epsilon)-Y_i(t_0))\int_{t_0}^{t_0+\epsilon}\Theta_j(u)du|{\cal
F}_{t_0}\right]+E\left[(Y_j(t_0+\epsilon)-Y_j(t_0))\int_{t_0}^{t_0+\epsilon}\Theta_i(u)du|{\cal
F}_{t_0}\right]\\
&=&I+I\!I+I\!I\!I+I\!V.
\end{eqnarray*}
\[
I=E[Y_i(t_0+\epsilon)Y_j(t_0+\epsilon)-Y_i(t_0)Y_j(t_0)|{\cal
F}_{t_0}]=E\left[\int_{t_0}^{t_0+\epsilon}\sigma_i(u)\sigma_j(u)\rho_{ij}(t)dt|{\cal
F}_{t_0}\right].
\]
By an argument similar to that involved in the proof of part (iii),
we conclude
$I=\sigma_i(t_0)\sigma_j(t_0)\rho_{ij}(t_0)\epsilon+o(\epsilon)$ and
\begin{eqnarray*}
I\!I&=&E\left[\int_{t_0}^{t_0+\epsilon}(\Theta_i(u)-\Theta_i(t_0))du\int_{t_0}^{t_0+\epsilon}\Theta_j(u)du|{\cal
F}_{t_0}\right]+\Theta_i(t_0)\epsilon
E\left[\int_{t_0}^{t_0+\epsilon}\Theta_j(u)du|{\cal
F}_{t_0}\right]\\
&=&o(\epsilon)+(M_i(\epsilon)-o(\epsilon))M_j(\epsilon)\\
&=&M_i(\epsilon)M_j(\epsilon)+o(\epsilon).
\end{eqnarray*}
By Cauchy's inequality under conditional expectation (note
$E[XY|{\cal F}]$ defines an inner product on $L^2(\Omega)$),
\begin{eqnarray*}
I\!I\!I &\le&
E\left[|Y_i(t_0+\epsilon)-Y_i(t_0)|\int_{t_0}^{t_0+\epsilon}|\Theta_j(u)|du|{\cal
F}_{t_0}\right]\\
&\le& M\epsilon\sqrt{E[(Y_i(t_0+\epsilon)-Y_i(t_0))^2|{\cal
F}_{t_0}]}\\
&\le& M\epsilon\sqrt{E[Y_i(t_0+\epsilon)^2-Y_i(t_0)^2|{\cal
F}_{t_0}]}\\
&\le&M\epsilon\sqrt{E[\int_{t_0}^{t_0+\epsilon}\Theta_i(u)^2du|{\cal
F}_{t_0}]}\\
&\le&M\epsilon\cdot M\sqrt{\epsilon}\\
&=&o(\epsilon)
\end{eqnarray*}
Similarly, $I\!V=o(\epsilon)$. In summary, we have
\[
E[(X_i(t_0+\epsilon)-X_i(t_0))(X_j(t_0+\epsilon)-X_j(t_0))|{\cal
F}_{t_0}]=M_i(\epsilon)M_j(\epsilon)+\sigma_i(t_0)\sigma_j(t_0)\rho_{ij}(t_0)\epsilon+o(\epsilon)+o(\epsilon).
\]This proves part (iv) and (v). Finally,
\[
\lim_{\epsilon\downarrow
0}\frac{C(\epsilon)}{\sqrt{V_1(\epsilon)V_2(\epsilon)}}=\lim_{\epsilon\downarrow
0}\frac{\rho(t_0)\sigma_1(t_0)\sigma_2(t_0)\epsilon+o(\epsilon)}{\sqrt{(\sigma_1^2(t_0)\epsilon+o(\epsilon))(\sigma_2^2(t_0)\epsilon+o(\epsilon))}}=
\rho(t_0).
\]This proves part (vi). Part (i) and (ii) are consequences of
general cases.

\end{proof}

\medskip

\noindent 4.18. (i)\begin{proof}
\[
d(e^{rt}\zeta_t)=(de^{-\theta W_t-\frac{1}{2}\theta^2t})=-e^{-\theta
W_t-\frac{1}{2}\theta^2t}\theta dW_t=-\theta (e^{rt}\zeta_t)dW_t,
\]where for the second ``=", we used the fact that $e^{-\theta
W_t-\frac{1}{2}\theta^2t}$ solves $dX_t=-\theta X_tdW_t$. Since
$d(e^{rt}\zeta_t)=re^{rt}\zeta_tdt+e^{rt}d\zeta_t$, we get
$d\zeta_t=-\theta \zeta_tdW_t-r\zeta_tdt$.
\end{proof}

(ii)\begin{proof}
\begin{eqnarray*}
d(\zeta_tX_t) &=&\zeta_tdX_t+X_td\zeta_t+dX_td\zeta_t\\
&=&\zeta_t(rX_tdt+\Delta_t(\alpha-r)S_tdt+\Delta_t\sigma
S_tdW_t)+X_t(-\theta\zeta_tdW_t-r\zeta_tdt)\\
& &+(rX_tdt+\Delta_t(\alpha-r)S_tdt+\Delta_t\sigma
S_tdW_t)(-\theta \zeta_tdW_t-r\zeta_tdt)\\
&=&\zeta_t(\Delta_t(\alpha-r)S_tdt+\Delta_t\sigma S_tdW_t)-\theta
X_t\zeta_tdW_t-\theta \Delta_t\sigma S_t\zeta_tdt\\
&=&\zeta_t\Delta_t\sigma S_tdW_t-\theta X_t\zeta_tdW_t.
\end{eqnarray*}So $\zeta_t X_t$ is a martingale.
\end{proof}

(iii)\begin{proof} By part (ii), $X_0=\zeta_0
X_0=E[\zeta_TX_t]=E[\zeta_TV_T]$. (This can be seen as a version of
risk-neutral pricing, only that the pricing is carried out under the
actual probability measure.)
\end{proof}

\medskip

\noindent 4.19. (i)\begin{proof} $B_t$ is a local martingale with
$[B]_t=\int_0^t sign(W_s)^2ds=t$. So by L\'{e}vy's theorem, $B_t$ is
a Brownian motion.
\end{proof}

(ii)\begin{proof} $d(B_tW_t)=B_tdW_t+sign(W_t)W_tdW_t+sign(W_t)dt$.
Integrate both sides of the resulting equation and the expectation,
we get
\[
E[B_tW_t]=\int_0^tE[sign(W_s)]ds=\int_0^tE[1_{\{W_s\ge
0\}}-1_{\{W_s<0\}}]ds=\frac{1}{2}t-\frac{1}{2}t=0.
\]
\end{proof}

(iii)\begin{proof} By It\^{o}'s formula, $dW_t^2=2W_tdW_t+dt$.
\end{proof}

(iv)\begin{proof}By It\^{o}'s formula,
\begin{eqnarray*}
d(B_tW_t^2)
&=&B_tdW_t^2+W_t^2dB_t+dB_tdW_t^2\\
&=&B_t(2W_tdW_t+dt)+W_t^2sign(W_t)dW_t+sign(W_t)dW_t(2W_tdW_t+dt)\\
&=&2B_tW_tdW_t+B_tdt+sign(W_t)W_t^2dW_t+2sign(W_t)W_tdt.
\end{eqnarray*}
So
\begin{eqnarray*}
E[B_tW_t^2]
&=&E[\int_0^tB_sds]+2E[\int_0^tsign(W_s)W_sds]\\
&=&\int_0^tE[B_s]ds+2\int_0^tE[sign(W_s)W_s]ds\\
&=&2\int_0^t(E[W_s1_{\{W_s\ge 0\}}]-E[W_s1_{\{W_s<0\}}])ds\\
&=&4\int_0^t\int_0^{\infty}x\frac{e^{-\frac{x^2}{2s}}}{\sqrt{2\pi
s}}dxds\\
&=&4\int_0^t\sqrt{\frac{s}{2\pi}}ds\\
&\ne& 0=E[B_t]\cdot E[W_t^2].
\end{eqnarray*}
Since $E[B_tW_t^2]\ne E[B_t]\cdot E[W_t^2]$, $B_t$ and $W_t$ are not
independent.
\end{proof}

\medskip

\noindent 4.20. (i)\begin{proof} $ f(x)=
\begin{cases}
x-K, &\mbox{if $x\ge K$}\\
0, &\mbox{if $x<K$}.
\end{cases}
$ So $ f'(x)=
\begin{cases}
1, &\mbox{if $x>K$}\\
\mbox{undefined}, &\mbox{if $x=K$}\\
0, &\mbox{if $x<K$}
\end{cases}
$ and $ f''(x)=
\begin{cases}
0, &\mbox{if $x\ne K$}\\
\mbox{undefined}, &\mbox{if $x=K$}.
\end{cases}
$
\end{proof}

(ii)\begin{proof}
$E[f(W_T)]=\int_K^{\infty}(x-K)\frac{e^{-\frac{x^2}{2T}}}{\sqrt{2\pi
T}}dx=\sqrt{\frac{T}{2\pi}}e^{-\frac{K^2}{2T}}-K\Phi(-\frac{K}{\sqrt{T}})$
where $\Phi$ is the distribution function of standard normal random
variable. If we suppose $\int_0^Tf''(W_t)dt=0$, the expectation of
RHS of (4.10.42) is equal to 0. So (4.10.42) cannot hold.
\end{proof}

(iii)\begin{proof}This is trivial to check.\end{proof}

(iv)\begin{proof}If $x=K$, $\lim_{n\to\infty}f_n(x)=\frac{1}{8n}=0$;
if $x>K$, for $n$ large enough, $x\ge K+\frac{1}{2n}$, so
$\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}(x-K)=x-K$; if $x<K$, for
$n$ large enough, $x\le K-\frac{1}{2n}$, so
$lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}0=0$. In summary,
$lim_{n\to\infty}f_n(x)=(x-K)^+$. Similarly, we can show
\begin{equation}
\lim_{n\to\infty}f_n'(x)=
\begin{cases}
0,&\mbox{if $x<K$}\\
\frac{1}{2},&\mbox{if $x=K$}\\
1,&\mbox{if $x>K$}.
\end{cases}
\end{equation}
\end{proof}

(v)\begin{proof}Fix $\omega$, so that $W_t(\omega)<K$ for any $t\in
[0,T]$. Since $W_t(\omega)$ can obtain its maximum on $[0,T]$, there
exists $n_0$, so that for any $n\ge n_0$, $\max_{0\le t\le
T}W_t(\omega)<K-\frac{1}{2n}$. So
\[
L_K(T)(\omega)=\lim_{n\to\infty}n\int_0^T1_{(K-\frac{1}{2n},K+\frac{1}{2n})}(W_t(\omega))dt=0.
\]
\end{proof}

(vi)\begin{proof}Take expectation on both sides of the formula
(4.10.45), we have
\[
E[L_K(T)]=E[(W_T-K)^+]>0.
\]So we cannot have $L_K(T)=0$ a.s..
\end{proof}

\begin{remark}
Cf. Kallenberg Theorem 22.5, or the paper by Alsmeyer and Jaeger (2005).
\end{remark}

\medskip

\noindent 4.21. (i) \begin{proof}There are two problems. First, the
transaction cost could be big due to active trading; second, the
purchases and sales cannot be made at exactly the same price $K$.
For more details, see Hull \cite{Hull00}.\end{proof}

(ii)\begin{proof}No. The RHS of (4.10.26) is a martingale, so its
expectation is 0. But $E[(S_T-K)^+]>0$. So $X_T\ne (S_T-K)^+$.
\end{proof}



\section{Risk-Neutral Pricing}

$\bigstar$ {\bf Comments}:

\medskip

1) {\it Heuristics to memorize the conditional Bayes formula (Lemma 5.2.2 on page 212 of the textbook)}
\[
\widetilde {\mathbb E}[Y | {\cal F}(s)] = \frac{1}{Z(s)} {\mathbb E} [YZ(t)|{\cal F}(s)].
\]
We recall an example in Durrett \cite[page~223]{Durrett10} (Example 5.1.3): suppose $\Omega_1$, $\Omega_2$, $\cdots$ is a finite or infinite partition of $\Omega$ into disjoint sets, each of which has positive probability, and let ${\cal F} = \sigma(\Omega_1, \Omega_2, \cdots)$ be the $\sigma$-field generated by these sets. Then
\[
{\mathbb E}[X|{\cal F}(s)] = \sum_i 1_{\Omega_i} \frac{{\mathbb E}[X;\Omega_i]}{{\mathbb P}(\Omega_i)}.
\]
Therefore, if ${\cal F}(s) = \sigma(\Omega_1, \Omega_2, \cdots)$, we have
\begin{eqnarray*}
\widetilde{\mathbb E}[Y|{\cal F}(s)] = \sum_i 1_{\Omega_i} \frac{\widetilde{\mathbb E}[Y;\Omega_i]}{\widetilde{\mathbb P}(\Omega_i)} = \sum_i 1_{\Omega_i} \frac{{\mathbb E}[YZ(t);\Omega_i]}{{\mathbb E}[Z(s); \Omega_i]}.
\end{eqnarray*}
Consequently, on any $\Omega_i$,
\[
\widetilde{\mathbb E}[Y|{\cal F}(s)] = \frac{{\mathbb E}[YZ(t);\Omega_i]/{\mathbb P}(\Omega_i)}{{\mathbb E}[Z(s); \Omega_i]/{\mathbb P}(\Omega_i)} = \frac{{\mathbb E}[YZ(t)|{\cal F}(s)]}{{\mathbb E}[Z(s)|{\cal F}(s)]} = \frac{1}{Z(s)} {\mathbb E} [YZ(t)|{\cal F}(s)].
\]

\medskip

2) {\it Heuristics to memorize the one-dimensional Girsanov's Theorem (Theorem 5.2.3 on page 212 of the textbook): suppose under a change of measure with density process $(Z(t))_{t\ge 0}$, $\widetilde W(t) = W(t) + \int_0^t \Theta(u)du$ becomes a martingale, then we must have}
\[
Z(t) = \exp \left\{-\int_0^t \Theta(u)dW(u) - \frac{1}{2}\int_0^t \Theta^2(u) du \right\}.
\]
Recall $Z$ is a positive martingale under the original probability measure ${\mathbb P}$, so by martingale representation theorem under the Brownian filtration, $Z$ must satisfy the SDE
\[
dZ(t) = Z(t) X(t) dW(t)
\]
for some adapted process $X(t)$.\footnote{The existence of $X$ is easy: if $dZ(t) = \xi(t)dW(t)$, then $X(t)=\frac{\xi(t)}{Z(t)}$.} Since an adapted process $M$ is a $\widetilde {\mathbb P}$-martingale if and only if $MZ$ is a ${\mathbb P}$-martingale,\footnote{To see this, note $\widetilde {\mathbb E}[M(t)|{\cal F}(s)] = {\mathbb E}[M(t)Z(t)|{\cal F}(s)]/Z(s)$.} we conclude $\widetilde W$ is a $\widetilde{\mathbb P}$-martingale if and only if $\widetilde W Z$ is a ${\mathbb P}$-martingale. Since
\begin{eqnarray*}
d[\widetilde W(t)Z(t)] &=& Z(t) [dW(t)+\Theta(t)dt] + \widetilde W(t)Z(t)X(t)dW(t) + [dW(t)+\Theta(t)dt] Z(t)X(t)dW(t) \\
&=& (\cdots)dW(t)+Z(t)[\Theta(t)+X(t)]dt,
\end{eqnarray*}
we must require $X(t)=-\Theta(t)$. So $Z(t) = \exp \left\{-\int_0^t \Theta(u)dW(u) - \frac{1}{2}\int_0^t \Theta^2(u) du \right\}$.

\medskip

3) {\it Main idea of Example 5.4.4.} Combining formula (5.4.15) and (5.4.22), we have
\[
d[D(t)X(t)] = \sum_{i=1}^m \Delta_i(t) d[D(t)S_i(t)] = \sum_{i=1}^m \Delta_i(t) D(t)S_i(t)[(\alpha_i(t)-R(t))dt+\sigma_i(t)dB_i(t)].
\]
To create arbitrage, we want $D(t)X(t)$ to have a deterministic and positive return rate, that is,
\[
\begin{cases}
\sum_{i=1}^m \Delta_i(t)S_i(t)\sigma_i(t) = 0 \\
\sum_{i=1}^m \Delta_i(t)S_i(t)[\alpha_i(t) - R(t)] > 0.
\end{cases}
\]
In Example 5.4.4, $i=2$, $\Delta_1(t) = \frac{1}{S_1(t) \sigma_1}$, $\Delta_2(t) = -\frac{1}{S_2(t)\sigma_2}$ and
\[
\frac{\alpha_1 - r}{\sigma_1} - \frac{\alpha - r}{\sigma_2} > 0.
\]

%\medskip
%
%4) {\it Comment on Black-Scholes formula for continuously paying dividend with constant coefficients.} We note the martingale property of discounted portfolio and the martingale property of discounted stock price are two separate issues. We need the former for risk-neutral pricing and the later for taking advantage of Black-Scholes formula.
%
%More precisely, the value of a self-financing portfolio $X(t)$ satisfies equation (5.5.2)
%\[
%dX(t) = R(t) X(t)dt + \Delta(t) S(t) \sigma(t) \left[\frac{\alpha(t)-R(t)}{\sigma(t)}dt+dW(t)\right].
%\]
%So the risk-neutral probability is the same as the case of a non-dividend-paying stock and the discounted portfolio value satisfies
%\[
%d[D(t)X(t)] = \Delta(t) D(t) S(t) \sigma(t) d\widetilde W(t).
%\]
%Therefore, the risk-neutral pricing formula for a European call option is the same as the case of a non-dividend-paying stock:
%\[
%V(t)=X(t) =\frac{1}{D(t)} \widetilde {\mathbb E}[D(T)(S(T)-K)^+|{\cal F}(t)].
%\]


\bigskip

\noindent 5.1. (i)\begin{proof}
\begin{eqnarray*}
df(X_t)&=&f'(X_t)dt+\frac{1}{2}f''(X_t)d\langle X\rangle_t\\
&=&f(X_t)(dX_t+\frac{1}{2}d\langle X\rangle_t)\\
&=&f(X_t)\left[\sigma_tdW_t+(\alpha_t-R_t-\frac{1}{2}\sigma_t^2)dt+\frac{1}{2}\sigma_t^2dt\right]\\
&=&f(X_t)(\alpha_t-R_t)dt+f(X_t)\sigma_tdW_t.
\end{eqnarray*}
 This is formula (5.2.20).
\end{proof}

(ii)\begin{proof}
$d(D_tS_t)=S_tdD_t+D_tdS_t+dD_tdS_t=-S_tR_tD_tdt+D_t\alpha_tS_tdt+D_t\sigma_tS_tdW_t=D_tS_t(\alpha_t-R_t)dt+D_tS_t\sigma_tdW_t$.
This is formula (5.2.20).
\end{proof}

\noindent 5.2. \begin{proof}By Lemma 5.2.2., $\widetilde
E[D_TV_T|{\cal F}_t]=E\left[\frac{D_TV_TZ_T}{Z_t}|{\cal
F}_t\right]$. Therefore (5.2.30) is equivalent to
$D_tV_tZ_t=E[D_TV_TZ_T|{\cal F}_t]$.
\end{proof}

\noindent 5.3. (i)\begin{proof}
\begin{eqnarray*}
c_x(0,x)&=&\frac{d}{dx}\widetilde E[e^{-rT}(xe^{\sigma \widetilde
W_T+(r-\frac{1}{2}\sigma^2)T}-K)^+]\\
&=&\widetilde E\left[e^{-rT}\frac{d}{dx}h(xe^{\sigma\widetilde
W_T+(r-\frac{1}{2}\sigma^2)T})\right]\\
&=&\widetilde E\left[e^{-rT}e^{\sigma \widetilde W_T+(r-\frac{1}{2}\sigma^2)T}1_{\{xe^{\sigma \widetilde W_T+(r-\frac{1}{2}\sigma^2)T}>K\}}\right]\\
&=&e^{-\frac{1}{2}\sigma^2T}\widetilde E\left[e^{\sigma \widetilde
W_T}1_{\{\widetilde
W_T>\frac{1}{\sigma}(\ln\frac{K}{x}-(r-\frac{1}{2}\sigma^2)T)\}}\right]\\
&=&e^{-\frac{1}{2}\sigma^2T}\widetilde
E\left[e^{\sigma\sqrt{T}\frac{\widetilde
W_T}{\sqrt{T}}}1_{\{\frac{\widetilde
W_T}{\sqrt{T}}-\sigma\sqrt{T}>\frac{1}{\sigma\sqrt{T}}(\ln\frac{K}{x}-(r-\frac{1}{2}\sigma^2)T)-\sigma\sqrt{T}\}}\right]\\
&=&e^{-\frac{1}{2}\sigma^2T}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}e^{\sigma\sqrt{T}z}1_{\{z-\sigma\sqrt{T}>-d_+(T,x)\}}dz\\
&=&\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{(z-\sigma\sqrt{T})^2}{2}}1_{\{z-\sigma\sqrt{T}>-d_+(T,x)\}}dz\\
&=&N(d_+(T,x)).
\end{eqnarray*}
\end{proof}

(ii)\begin{proof}If we set $\widehat Z_T=e^{\sigma \widetilde
W_T-\frac{1}{2}\sigma^2T}$ and $\widehat Z_t=\widetilde E[\widehat
Z_T|{\cal F}_t]$, then $\widehat Z$ is a $\widetilde P$-martingale,
$\widehat Z_t>0$ and $E[\widehat Z_T]=\widetilde E[e^{\sigma
\widetilde W_T-\frac{1}{2}\sigma^2T}]=1$. So if we define $\widehat
P$ by $d\widehat P=Z_Td\widetilde P$ on ${\cal F}_T$, then $\widehat
P$ is a probability measure equivalent to $\widetilde P$, and
\[
c_x(0,x)=\widetilde E[\widehat Z_T1_{\{S_T>K\}}]=\widehat P(S_T>K).
\]
Moreover, by Girsanov's Theorem, $\widehat W_t=\widetilde
W_t+\int_0^t(-\sigma)du=\widetilde W_t-\sigma t$ is a $\widehat
P$-Brownian motion (set $\Theta=-\sigma$ in Theorem 5.4.1.)
\end{proof}

(iii)\begin{proof}$S_T=xe^{\sigma\widetilde
W_T+(r-\frac{1}{2}\sigma^2)T}=xe^{\sigma\widehat
W_T+(r+\frac{1}{2}\sigma^2)T}$. So
\begin{eqnarray*}
\widehat P(S_T>K)=\widehat P(xe^{\sigma\widehat
W_T+(r+\frac{1}{2}\sigma^2)T}>K)=\widehat P\left(\frac{\widehat
W_T}{\sqrt{T}}>-d_+(T,x)\right)=N(d_+(T,x)).
\end{eqnarray*}
\end{proof}

\noindent 5.4. First, a few typos. In the SDE for $S$,
$``\sigma(t)d\widetilde W(t)" \to ``\sigma(t)S(t)d\widetilde W(t)"$.
In the first equation for $c(0, S(0))$, $E\to \widetilde E$. In the
second equation for $c(0,S(0))$, the variable for BSM should be
\[BSM\left(T,S(0);K,\frac{1}{T}\int_0^Tr(t)dt,\sqrt{\frac{1}{T}\int_0^T\sigma^2(t)dt}\right).\]

(i)\begin{proof} $d\ln S_t=\frac{dS_t}{S_t}-\frac{1}{2S_t^2}d\langle
S\rangle_t=r_tdt+\sigma_td\widetilde W_t-\frac{1}{2}\sigma_t^2dt$.
So
$S_T=S_0\exp\{\int_0^T(r_t-\frac{1}{2}\sigma_t^2)dt+\int_0^T\sigma_td\widetilde
W_t\}$. Let
$X=\int_0^T(r_t-\frac{1}{2}\sigma_t^2)dt+\int_0^T\sigma_td\widetilde
W_t$. The first term in the expression of $X$ is a number and the
second term is a Gaussian random variable
$N(0,\int_0^T\sigma_t^2dt)$, since both $r$ and $\sigma$ ar
deterministic. Therefore, $S_T=S_0e^X$, with $X\thicksim
N(\int_0^T(r_t-\frac{1}{2}\sigma_t^2)dt,\int_0^T\sigma_t^2dt)$,.
\end{proof}

(ii)\begin{proof} For the standard BSM model with constant
volatility $\Sigma$ and interest rate $R$, under the risk-neutral
measure, we have $S_T=S_0e^Y$, where
$Y=(R-\frac{1}{2}\Sigma^2)T+\Sigma \widetilde W_T\thicksim
N((R-\frac{1}{2}\Sigma^2)T,\Sigma^2T)$, and $\widetilde
E[(S_0e^Y-K)^+]=e^{RT}BSM(T,S_0;K,R,\Sigma)$. Note
$R=\frac{1}{T}(E[Y]+\frac{1}{2}Var(Y))$ and
$\Sigma=\sqrt{\frac{1}{T}Var(Y)}$, we can get
\[
\widetilde
E[(S_0e^Y-K)^+]=e^{E[Y]+\frac{1}{2}Var(Y)}BSM\left(T,S_0;K,\frac{1}{T}\left(E[Y]+\frac{1}{2}Var(Y)\right),\sqrt{\frac{1}{T}Var(Y)}\right).
\]
So for the model in this problem,
\begin{eqnarray*}
c(0,S_0)&=&e^{-\int_0^Tr_tdt}\widetilde E[(S_0e^X-K)^+]\\
&=&e^{-\int_0^Tr_tdt}e^{E[X]+\frac{1}{2}Var(X)}BSM\left(T,S_0;K,\frac{1}{T}\left(E[X]+\frac{1}{2}Var(X)\right),\sqrt{\frac{1}{T}Var(X)}\right)\\
&=&BSM\left(T,S_0;K,\frac{1}{T}\int_0^Tr_tdt,\sqrt{\frac{1}{T}\int_0^T\sigma_t^2dt}\right).
\end{eqnarray*}
\end{proof}

\noindent 5.5. (i)\begin{proof} Let $f(x)=\frac{1}{x}$, then
$f'(x)=-\frac{1}{x^2}$ and $f''(x)=\frac{2}{x^3}$. Note
$dZ_t=-Z_t\Theta_tdW_t$, so
\[
d\left(\frac{1}{Z_t}\right)=f'(Z_t)dZ_t+\frac{1}{2}f''(Z_t)dZ_tdZ_t=-\frac{1}{Z_t^2}(-Z_t)\Theta_tdW_t+\frac{1}{2}\frac{2}{Z_t^3}Z_t^2\Theta_t^2dt=\frac{\Theta_t}{Z_t}dW_t+\frac{\Theta_t^2}{Z_t}dt.
\]
\end{proof}

(ii)\begin{proof} By Lemma 5.2.2., for $s,t\ge 0$ with $s<t$,
$\widetilde M_s=\widetilde E[\widetilde M_t|{\cal
F}_s]=E\left[\frac{Z_t\widetilde M_t}{Z_s}|{\cal F}_s\right]$. That
is, $E[Z_t\widetilde M_t|{\cal F}_s]=Z_s\widetilde M_s$. So
$M=Z\widetilde M$ is a $P$-martingale.
\end{proof}

(iii)\begin{proof}
\[
d\widetilde M_t=d\left(M_t\cdot
\frac{1}{Z_t}\right)=\frac{1}{Z_t}dM_t+M_td\frac{1}{Z_t}+dM_td\frac{1}{Z_t}=\frac{\Gamma_t}{Z_t}dW_t+\frac{M_t\Theta_t}{Z_t}dW_t+\frac{M_t\Theta_t^2}{Z_t}dt+\frac{\Gamma_t\Theta_t}{Z_t}dt.
\]
\end{proof}

(iv)\begin{proof}In part (iii), we have
\[
d\widetilde
M_t=\frac{\Gamma_t}{Z_t}dW_t+\frac{M_t\Theta_t}{Z_t}dW_t+\frac{M_t\Theta_t^2}{Z_t}dt+\frac{\Gamma_t\Theta_t}{Z_t}dt=\frac{\Gamma_t}{Z_t}(dW_t+\Theta_tdt)
+\frac{M_t\Theta_t}{Z_t}(dW_t+\Theta_tdt).
\]
Let $\widetilde \Gamma_t=\frac{\Gamma_t+M_t\Theta_t}{Z_t}$, then
$d\widetilde M_t=\widetilde \Gamma_td\widetilde W_t$. This proves
Corollary 5.3.2.
\end{proof}

 \noindent 5.6. \begin{proof} By Theorem 4.6.5, it
suffices to show $\widetilde W_i(t)$ is an ${\cal F}_t$-martingale
under $\widetilde P$ and $[\widetilde W_i,\widetilde
W_j](t)=t\delta_{ij}$ $(i,j=1,2)$. Indeed, for $i=1,2$, $\widetilde
W_i(t)$ is an ${\cal F}_t$-martingale under $\widetilde P$ if and
only if $\widetilde W_i(t)Z_t$ is an ${\cal F}_t$-martingale under
$P$, since
\[
\widetilde E[\widetilde W_i(t)|{\cal F}_s]=E\left[\frac{\widetilde
W_i(t)Z_t}{Z_s}|{\cal F}_s\right].
\]
By It\^{o}'s product formula, we have
\begin{eqnarray*}
d(\widetilde W_i(t)Z_t)&=&\widetilde W_i(t)dZ_t+Z_td\widetilde
W_i(t)+dZ_td\widetilde W_i(t)\\
&=&\widetilde
W_i(t)(-Z_t)\Theta(t)\cdot dW_t+Z_t(dW_i(t)+\Theta_i(t)dt)+(-Z_t\Theta_t\cdot dW_t)(dW_i(t)+\Theta_i(t)dt)\\
&=&\widetilde
W_i(t)(-Z_t)\sum_{j=1}^d\Theta_j(t)dW_j(t)+Z_t(dW_i(t)+\Theta_i(t)dt)-Z_t\Theta_i(t)dt\\
&=&\widetilde
W_i(t)(-Z_t)\sum_{j=1}^d\Theta_j(t)dW_j(t)+Z_tdW_i(t)\\
\end{eqnarray*}
This shows $\widetilde W_i(t)Z_t$ is an ${\cal F}_t$-martingale
under $P$. So $\widetilde W_i(t)$ is an ${\cal F}_t$-martingale
under $\widetilde P$. Moreover,
\[
[\widetilde W_i,\widetilde
W_j](t)=\left[W_i+\int_0^{\cdot}\Theta_i(s)ds,W_j+\int_0^{\cdot}\Theta_j(s)ds\right](t)=[W_i,W_j](t)=t\delta_{ij}.
\]
Combined, this proves the two-dimensional Girsanov's Theorem.
\end{proof}


\noindent 5.7. (i)\begin{proof} Let $a$ be any strictly positive
number. We define $X_2(t)=(a+X_1(t))D(t)^{-1}$. Then \[
P\left(X_2(T)\ge \frac{X_2(0)}{D(T)}\right)=P(a+X_1(T)\ge
a)=P(X_1(T)\ge 0)=1,\]
 and $P\left(X_2(T)>\frac{X_2(0)}{D(T)}\right)=P(X_1(T)>0)>0$, since
 $a$ is arbitrary, we have proved the claim of this problem.
\end{proof}

 \begin{remark} The intuition is that we invest the positive
 starting fund $a$ into the money market account, and construct
 portfolio $X_1$ from zero cost. Their sum should be able to beat
 the return of money market account.
 \end{remark}


(ii)\begin{proof} We define $X_1(t)=X_2(t)D(t)-X_2(0)$. Then
$X_1(0)=0$,
\begin{eqnarray*}
P(X_1(T)\ge 0)=P\left(X_2(T)\ge \frac{X_2(0)}{D(T)}\right)=1,\;
P(X_1(T)>0)=P\left(X_2(T)>\frac{X_2(0)}{D(T)}\right)>0.
\end{eqnarray*}
\end{proof}

\noindent 5.8. The basic idea is that for any positive $\widetilde
P$-martingale $M$, $dM_t=M_t\cdot\frac{1}{M_t}dM_t$. By Martingale
Representation Theorem, $dM_t=\widetilde \Gamma_td\widetilde W_t$
for some adapted process $\widetilde \Gamma_t$. So
$dM_t=M_t(\frac{\widetilde \Gamma_t}{M_t})d\widetilde W_t$, i.e. any
positive martingale must be the exponential of an integral w.r.t.
Brownian motion. Taking into account discounting factor and apply
It\^{o}'s product rule, we can show every strictly positive asset is
a generalized geometric Brownian motion.

(i)\begin{proof} $V_tD_t=\widetilde E[e^{-\int_0^TR_udu}V_T|{\cal
F}_t]=\widetilde E[D_TV_T|{\cal F}_t]$. So $(D_tV_t)_{t\ge 0}$ is a
$\widetilde P$-martingale. By Martingale Representation Theorem,
there exists an adapted process $\widetilde \Gamma_t$, $0\le t\le
T$, such that $D_tV_t=\int_0^t\widetilde \Gamma_sd\widetilde W_s$,
or equivalently, $V_t=D_t^{-1}\int_0^t\widetilde\Gamma_sd\widetilde
W_s$. Differentiate both sides of the equation, we get
$dV_t=R_tD_t^{-1}\int_0^t\widetilde \Gamma_s d\widetilde W_s
dt+D_t^{-1}\widetilde \Gamma_td\widetilde W_t$, i.e.
$dV_t=R_tV_tdt+\frac{\widetilde \Gamma_t}{D_t}dW_t$.
\end{proof}

(ii)\begin{proof} We prove the following more general lemma.

\begin{lemma} Let $X$ be an almost surely positive random variable
(i.e. $X>0$ a.s.) defined on the probability space $(\Omega,{\cal
G}, P)$. Let ${\cal F}$ be a sub $\sigma$-algebra of ${\cal G}$,
then $Y=E[X|{\cal F}]>0$ a.s.
\end{lemma}
\begin{proof}
By the property of conditional expectation $Y_t\ge 0$ a.s. Let
$A=\{Y=0\}$, we shall show $P(A)=0$. Indeed, note $A\in {\cal F}$,
$0=E[YI_A]=E[E[X|{\cal F}]I_A]=E[XI_A]=E[X1_{A\cap\{X\ge
1\}}]+\sum_{n=1}^{\infty}E[X1_{A\cap\{\frac{1}{n}>X\ge
\frac{1}{n+1}\}}]\ge P(A\cap \{X\ge
1\})+\sum_{n=1}^{\infty}\frac{1}{n+1}P(A\cap
\{\frac{1}{n}>X\ge\frac{1}{n+1}\})$. So $P(A\cap\{X\ge 1\})=0$ and
$P(A\cap\{\frac{1}{n}>X\ge\frac{1}{n+1}\})=0$, $\forall n\ge 1$.
This in turn implies $P(A)=P(A\cap\{X>0\})=P(A\cap\{X\ge
1\})+\sum_{n=1}^{\infty}P(A\cap\{\frac{1}{n}>X\ge\frac{1}{n+1}\})=0$.
\end{proof}

By the above lemma, it is clear that for each $t\in [0,T]$,
$V_t=\widetilde E[e^{-\int_t^TR_udu}V_T|{\cal F}_t]>0$ a.s..
Moreover, by a classical result of martingale theory (Revuz and Yor
\cite{RY98}, Chapter II, Proposition (3.4)), we have the following
stronger result: for a.s. $\omega$, $V_t(\omega)>0$ for any $t\in
[0,T]$.
\end{proof}

(iii)\begin{proof} By (ii), $V>0$ a.s., so
$dV_t=V_t\frac{1}{V_t}dV_t=V_t\frac{1}{V_t}\left(R_tV_tdt+\frac{\widetilde
\Gamma_t}{D_t}d\widetilde W_t\right)=V_tR_tdt+V_t\frac{\widetilde
\Gamma_t}{V_tD_t}d\widetilde W_t=R_tV_tdt+\sigma_tV_td\widetilde
W_t$, where $\sigma_t=\frac{\widetilde \Gamma_t}{V_tD_t}$. This
shows $V$ follows a generalized geometric Brownian motion.
\end{proof}

\noindent 5.9. \begin{proof} $c(0,T,x,K)=xN(d_+)-Ke^{-rT}N(d_-)$
with
$d_{\pm}=\frac{1}{\sigma\sqrt{T}}(\ln\frac{x}{K}+(r\pm\frac{1}{2}\sigma^2)T)$.
Let $f(y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}$, then
$f'(y)=-yf(y)$,
\begin{eqnarray*}
c_K(0,T,x,K)&=& xf(d_+)\frac{\partial d_+}{\partial
y}-e^{-rT}N(d_-)-Ke^{-rT}f(d_-)\frac{\partial d_-}{\partial
y}\\
&=&xf(d_+)\frac{-1}{\sigma\sqrt{T}K}-e^{-rT}N(d_-)+e^{-rT}f(d_-)\frac{1}{\sigma\sqrt{T}},
\end{eqnarray*} and
\begin{eqnarray*}
& &c_{KK}(0,T,x,K)\\
&=&xf(d_+)\frac{1}{\sigma\sqrt{T}K^2}-\frac{x}{\sigma\sqrt{T}K}f(d_+)(-d_+)\frac{\partial
d_+}{\partial y}-e^{-rT}f(d_-)\frac{\partial d_-}{\partial
y}+\frac{e^{-rT}}{\sigma\sqrt{T}}(-d_-)f(d_-)\frac{d_-}{\partial
y}\\
&=&\frac{x}{\sigma\sqrt{T}K^2}f(d_+)+\frac{xd_+}{\sigma\sqrt{T}K}f(d_+)\frac{-1}{K\sigma\sqrt{T}}-e^{-rT}f(d_-)\frac{-1}{K\sigma\sqrt{T}}-\frac{e^{-rT}d_-}{\sigma\sqrt{T}}f(d_-)\frac{-1}{K\sigma\sqrt{T}}\\
&=&f(d_+)\frac{x}{K^2\sigma\sqrt{T}}[1-\frac{d_+}{\sigma\sqrt{T}}]+\frac{e^{-rT}f(d_-)}{K\sigma\sqrt{T}}[1+\frac{d_-}{\sigma\sqrt{T}}]\\
&=&\frac{e^{-rT}}{K\sigma^2T}f(d_-)d_+-\frac{x}{K^2\sigma^2T}f(d_+)d_-.
\end{eqnarray*}
\end{proof}

\noindent 5.10. (i)\begin{proof} At time $t_0$, the value of the
chooser option is
$V(t_0)=\max\{C(t_0),P(t_0)\}=\max\{C(t_0),C(t_0)-F(t_0)\}=C(t_0)+\max\{0,-F(t_0)\}=C(t_0)+(e^{-r(T-t_0)}K-S(t_0))^+.$
\end{proof}

(ii)\begin{proof}By the risk-neutral pricing formula,
$V(0)=\widetilde E[e^{-rt_0}V(t_0)]=\widetilde
E[e^{-rt_0}C(t_0)+(e^{-rT}K-e^{-rt_0}S(t_0)^+]=C(0)+\widetilde
E[e^{-rt_0}(e^{-r(T-t_0)}K-S(t_0))^+]$. The first term is the value
of a call expiring at time $T$ with strike price $K$ and the second
term is the value of a put expiring at time $t_0$ with strike price
$e^{-r(T-t_0)}K$.

\end{proof}


\noindent 5.11. \begin{proof} We first make an analysis which leads
to the hint, then we give a formal proof.

(Analysis) If we want to construct a portfolio $X$ that exactly
replicates the cash flow, we must find a solution to the backward
SDE
\[
\begin{cases}
dX_t=\Delta_tdS_t+R_t(X_t-\Delta_tS_t)dt-C_tdt\\
X_T=0.
\end{cases}
\]
Multiply $D_t$ on both sides of the first equation and apply
It\^{o}'s product rule, we get
$d(D_tX_t)=\Delta_td(D_tS_t)-C_tD_tdt$. Integrate from $0$ to $T$,
we have $D_TX_T-D_0X_0=\int_0^T\Delta_td(D_tS_t)-\int_0^TC_tD_tdt$.
By the terminal condition, we get
$X_0=D_0^{-1}(\int_0^TC_tD_tdt-\int_0^T\Delta_td(D_tS_t))$. $X_0$ is
the theoretical, no-arbitrage price of the cash flow, provided we
can find a trading strategy $\Delta$ that solves the BSDE. Note the
SDE for $S$ gives $d(D_tS_t)=(D_tS_t)\sigma_t(\theta_tdt+dW_t)$,
where $\theta_t=\frac{\alpha_t-R_t}{\sigma_t}$. Take the proper
change of measure so that $\widetilde W_t=\int_0^t\theta_sds+W_t$ is
a Brownian motion under the new measure $\widetilde P$, we get
\[
\int_0^TC_tD_tdt=D_0X_0+\int_0^T\Delta_td(D_tS_t)=D_0X_0+\int_0^T\Delta_t(D_tS_t)\sigma_td\widetilde
W_t.
\]This  says the random variable $\int_0^TC_tD_tdt$ has a
stochastic integral representation
$D_0X_0+\int_0^T\Delta_tD_tS_t\sigma_td\widetilde W_t$. This
inspires us to consider the martingale generated by
$\int_0^TC_tD_tdt$, so that we can apply Martingale Representation
Theorem and get a formula for $\Delta$ by comparison of the
integrands.

(Formal proof) Let $M_T=\int_0^TC_tD_tdt$, and $M_t=\widetilde
E[M_T|{\cal F}_t]$. Then by Martingale Representation Theorem, we
can find an adapted process $\widetilde \Gamma_t$, so that
$M_t=M_0+\int_0^t\widetilde \Gamma_td\widetilde W_t$. If we set
$\Delta_t=\frac{\widetilde \Gamma_t}{D_tS_t\sigma_t}$, we can check
$X_t=D_t^{-1}(D_0X_0+\int_0^t\Delta_ud(D_uS_u)-\int_0^tC_uD_udu)$,
with $X_0=M_0=\widetilde E[\int_0^TC_tD_tdt]$ solves the SDE
\[
\begin{cases}
dX_t=\Delta_tdS_t+R_t(X_t-\Delta_tS_t)dt-C_tdt\\
X_T=0.
\end{cases}
\]Indeed, it is easy to see that $X$ satisfies the first equation.
To  check the terminal condition, we note
$X_TD_T=D_0X_0+\int_0^T\Delta_tD_tS_t\sigma_td\widetilde
W_t-\int_0^TC_tD_tdt=M_0+\int_0^T\widetilde \Gamma_td\widetilde
W_t-M_T=0$. So $X_T=0$. Thus, we have found a trading strategy
$\Delta$, so that the corresponding portfolio $X$ replicates the
cash flow and has zero terminal value. So $X_0=\widetilde
E[\int_0^TC_tD_tdt]$ is the no-arbitrage price of the cash flow at
time zero.
\end{proof}

\begin{remark} As shown in the analysis,
$d(D_tX_t)=\Delta_td(D_tS_t)-C_tD_tdt$. Integrate from $t$ to $T$,
we get $0-D_tX_t=\int_t^T\Delta_ud(D_uS_u)-\int_t^TC_uD_udu$. Take
conditional expectation w.r.t. ${\cal F}_t$ on both sides, we get
$-D_tX_t=-\widetilde E[\int_t^TC_uD_udu|{\cal F}_t]$. So
$X_t=D_t^{-1}\widetilde E[\int_t^TC_uD_udu|{\cal F}_t]$. This is the
no-arbitrage price of the cash flow at time $t$, and we have
justified formula (5.6.10) in the textbook.
\end{remark}



\noindent 5.12. (i)\begin{proof} $d\widetilde
B_i(t)=dB_i(t)+\gamma_i(t)dt=\sum_{j=1}^d\frac{\sigma_{ij}(t)}{\sigma_i(t)}dW_j(t)+\sum_{j=1}^d\frac{\sigma_{ij}(t)}{\sigma_i(t)}\Theta_j(t)dt
=\sum_{j=1}^d\frac{\sigma_{ij}(t)}{\sigma_i(t)}d\widetilde W_j(t)$.
So $B_i$ is a martingale. Since $d\widetilde B_i(t)d\widetilde
B_i(t)=\sum_{j=1}^d\frac{\sigma_{ij}(t)^2}{\sigma_i(t)^2}dt=dt$, by
L\'{e}vy's Theorem, $\widetilde B_i$ is a Brownian motion under
$\widetilde P$.\end{proof}

(ii)\begin{proof}
\begin{eqnarray*}
dS_i(t)&=&R(t)S_i(t)dt+\sigma_i(t)S_i(t)d\widetilde
B_i(t)+(\alpha_i(t)-R(t))S_i(t)dt-\sigma_i(t)S_i(t)\gamma_i(t)dt\\
&=&R(t)S_i(t)dt+\sigma_i(t)S_i(t)d\widetilde
B_i(t)+\sum_{j=1}^d\sigma_{ij}(t)\Theta_j(t)S_i(t)dt-S_i(t)\sum_{j=1}^d\sigma_{ij}(t)\Theta_j(t)dt\\
&=&R(t)S_i(t)dt+\sigma_i(t)S_i(t)d\widetilde B_i(t).
\end{eqnarray*}

\end{proof}


(iii)\begin{proof} $d\widetilde B_i(t)d\widetilde
B_k(t)=(dB_i(t)+\gamma_i(t)dt)(dB_j(t)+\gamma_j(t)dt)=dB_i(t)dB_j(t)=\rho_{ik}(t)dt.$
\end{proof}

(iv)\begin{proof} By It\^{o}'s product rule and martingale property,
\begin{eqnarray*}
E[B_i(t)B_k(t)]&=&E[\int_0^tB_i(s)dB_k(s)]+E[\int_0^tB_k(s)dB_i(s)]+E[\int_0^tdB_i(s)dB_k(s)]\\
&=&E[\int_0^t\rho_{ik}(s)ds]=\int_0^t\rho_{ik}(s)ds.
\end{eqnarray*}
Similarly, by part (iii), we can show $\widetilde E[\widetilde
B_i(t)\widetilde B_k(t)]=\int_0^t\rho_{ik}(s)ds.$
\end{proof}

(v)\begin{proof} By It\^{o}'s product formula,
\[
E[B_1(t)B_2(t)]=E[\int_0^tsign(W_1(u))du]=\int_0^t[P(W_1(u)\ge
0)-P(W_1(u)<0)]du=0.
\]Meanwhile,
\begin{eqnarray*}
\widetilde E[\widetilde B_1(t)\widetilde B_2(t)]&=&\widetilde
E[\int_0^tsign(W_1(u))du\\
&=&\int_0^t[\widetilde P(W_1(u)\ge 0)-\widetilde
P(W_1(u)<0)]du\\
&=&\int_0^t[\widetilde P(\widetilde
W_1(u)\ge u)-\widetilde P(\widetilde W_1(u)<u)]du\\
&=&\int_0^t2\left(\frac{1}{2}-\widetilde P(\widetilde W_1(u)<u)\right)du\\
&<&0,
\end{eqnarray*}for any $t>0$. So $E[B_1(t)B_2(t)]=\widetilde E[\widetilde B_1(t)\widetilde
B_2(t)]$ for all $t>0$.
\end{proof}

\noindent 5.13. (i)\begin{proof} $\widetilde E[W_1(t)]=\widetilde
E[\widetilde W_1(t)]=0$ and $\widetilde E[W_2(t)]=\widetilde
E[\widetilde W_2(t)-\int_0^t\widetilde W_1(u)du]=0$, for all $t\in
[0,T]$.
\end{proof}

(ii)\begin{proof} \begin{eqnarray*} \widetilde Cov[W_1(T),W_2(T)]
&=&\widetilde E[W_1(T)W_2(T)]\\
&=&\widetilde
E\left[\int_0^TW_1(t)dW_2(t)+\int_0^TW_2(t)dW_1(t)\right]\\
&=&\widetilde E\left[\int_0^T\widetilde W_1(t)(d\widetilde
W_2(t)-\widetilde W_1(t)dt)\right]+\widetilde
E\left[\int_0^TW_2(t)d\widetilde
W_1(t)\right]\\
&=&-\widetilde E\left[\int_0^T\widetilde
W_1(t)^2dt\right]\\
&=&-\int_0^Ttdt\\
&=&-\frac{1}{2}T^2.
\end{eqnarray*}
\end{proof}

\noindent 5.14. Equation (5.9.6) can be transformed into
$d(e^{-rt}X_t)=\Delta_t[d(e^{-rt}S_t)-ae^{-rt}dt]=\Delta_te^{-rt}[dS_t-rS_tdt-adt]$.
So, to make the discounted portfolio value $e^{-rt}X_t$ a
martingale, we are motivated to change the measure in such a way
that $S_t-r\int_0^tS_udu-at$ is a martingale under the new measure.
To do this, we note the SDE for $S$ is $dS_t=\alpha_tS_tdt+\sigma
S_tdW_t$. Hence $dS_t-rS_tdt-adt=[(\alpha_t-r)S_t-a]dt+\sigma
S_tdW_t=\sigma S_t\left[\frac{(\alpha_t-r)S_t-a}{\sigma
S_t}dt+dW_t\right]$. Set $\theta_t=\frac{(\alpha_t-r)S_t-a}{\sigma
S_t}$ and $\widetilde W_t=\int_0^t\theta_sds+W_t$, we can find an
equivalent probability measure $\widetilde P$, under which $S$
satisfies the SDE $dS_t=rS_tdt+\sigma S_td\widetilde W_t+adt$ and
$\widetilde W_t$ is a BM. This is the rational for formula (5.9.7).

This is a good place to pause and think about the meaning of
``martingale measure." {\it What is to be a martingale?} The new
measure $\widetilde P$ should be such that {\it the discounted value
process of the replicating portfolio} is a martingale, {\it not the
discounted price process of the underlying}. First, we want $D_tX_t$
to be a martingale under $\widetilde P$ because we suppose that $X$
is able to replicate the derivative payoff at terminal time,
$X_T=V_T$. In order to avoid arbitrage, we must have $X_t=V_t$ for
any $t\in [0,T]$. The difficulty is how to calculate $X_t$ and the
magic is brought by the martingale measure in the following line of
reasoning: $V_t=X_t=D_t^{-1}\widetilde E[D_TX_T|{\cal
F}_t]=D_t^{-1}\widetilde E[D_TV_T|{\cal F}_t]$. You can think of
martingale measure as a calculational convenience. That is {\it all} about
martingale measure! {\it Risk neutral} is just a {\it perception}, referring to the actual effect of constructing a hedging portfolio!
 Second, we note
when the portfolio is self-financing, the discounted price process
of the underlying is a martingale under $\widetilde P$, as in the
classical Black-Scholes-Merton model without dividends or cost of
carry. This is not a coincidence. Indeed, we have in this case the
relation $d(D_tX_t)=\Delta_td(D_tS_t)$. So $D_tX_t$ being a
martingale under $\widetilde P$ is more or less equivalent to
$D_tS_t$ being a martingale under $\widetilde P$.   However, when
the underlying pays dividends, or there is cost of carry,
$d(D_tX_t)=\Delta_td(D_tS_t)$ no longer holds, as shown in formula
(5.9.6). The portfolio is no longer {\it self-financing}, but {\it
self-financing with consumption}. What we still want to retain is
the martingale property of $D_tX_t$, not that of $D_tS_t$. This is
how we choose martingale measure in the above paragraph.

Let $V_T$ be a payoff at time $T$, then for the martingale
$M_t=\widetilde E[e^{-rT}V_T|{\cal F}_t]$, by Martingale
Representation Theorem, we can find an adapted process $\widetilde
\Gamma_t$, so that $M_t=M_0+\int_0^t\widetilde \Gamma_sd\widetilde
W_s$. If we let $\Delta_t=\frac{\widetilde \Gamma_te^{rt}}{\sigma
S_t}$, then the value of the corresponding portfolio $X$ satisfies
$d(e^{-rt}X_t)=\widetilde \Gamma_td\widetilde W_t$. So by setting
$X_0=M_0=\widetilde E[e^{-rT}V_T]$, we must have $e^{-rt}X_t=M_t$,
for all $t\in [0,T]$. In particular, $X_T=V_T$. Thus the portfolio
perfectly hedges $V_T$. This justifies the risk-neutral pricing of
European-type contingent claims in the model where cost of carry
exists. Also note the risk-neutral measure is different from the one
in case of no cost of carry.

Another perspective for perfect replication is the following. We
need to solve the backward SDE
\begin{eqnarray*}
\begin{cases}
dX_t=\Delta_tdS_t-a\Delta_tdt+r(X_t-\Delta_tS_t)dt\\
X_T=V_T
\end{cases}
\end{eqnarray*}
for two unknowns, $X$ and $\Delta$. To do so, we find a probability
measure $\widetilde P$, under which $e^{-rt}X_t$ is a martingale,
then $e^{-rt}X_t=\widetilde E[e^{-rT}V_T|{\cal F}_t]:=M_t$.
Martingale Representation Theorem gives
$M_t=M_0+\int_0^t\widetilde\Gamma_ud\widetilde W_u$ for some adapted
process $\widetilde \Gamma$. This would give us a theoretical
representation of $\Delta$ by comparison of integrands, hence a
perfect replication of $V_T$.

(i)\begin{proof} As indicated in the above analysis, if we have
(5.9.7) under $\widetilde P$, then
$d(e^{-rt}X_t)=\Delta_t[d(e^{-rt}S_t)-ae^{-rt}dt]=\Delta_te^{-rt}\sigma
S_td\widetilde W_t$. So $(e^{-rt}X_t)_{t\ge 0}$, where $X$ is given
by (5.9.6), is a $\widetilde P$-martingale.


\end{proof}

(ii)\begin{proof} By It\^{o}'s formula, $dY_t=Y_t[\sigma d\widetilde
W_t+(r-\frac{1}{2}\sigma^2)dt]+\frac{1}{2}Y_t\sigma^2dt=Y_t(\sigma
d\widetilde W_t+rdt)$. So $d(e^{-rt}Y_t)=\sigma
e^{-rt}Y_td\widetilde W_t$ and $e^{-rt}Y_t$ is a $\widetilde
P$-martingale. Moreover, if $S_t=S_0Y_t+Y_t\int_0^t\frac{a}{Y_s}ds$,
then
\[
dS_t=S_0dY_t+\int_0^t\frac{a}{Y_s}dsdY_t+adt=\left(S_0+\int_0^t\frac{a}{Y_s}ds\right)Y_t(\sigma
d\widetilde W_t+rdt)+adt=S_t(\sigma d\widetilde W_t+rdt)+adt.
\]This shows $S$ satisfies (5.9.7).
\end{proof}

\begin{remark} To obtain this formula for $S$, we first set
$U_t=e^{-rt}S_t$ to remove the $rS_tdt$ term. The SDE for $U$ is
$dU_t=\sigma U_td\widetilde W_t+ae^{-rt}dt$. Just like solving
linear ODE, to remove $U$ in the $d\widetilde W_t$ term, we consider
$V_t=U_te^{-\sigma \widetilde W_t}$. It\^{o}'s product formula
yields \begin{eqnarray*} dV_t&=&e^{-\sigma \widetilde
W_t}dU_t+U_te^{-\sigma \widetilde W_t}\left((-\sigma)d\widetilde
W_t+\frac{1}{2}\sigma^2dt\right)+dU_t\cdot e^{-\sigma \widetilde
W_t}\left((-\sigma)d\widetilde
W_t+\frac{1}{2}\sigma^2dt\right)\\
&=&e^{-\sigma \widetilde W_t}ae^{-rt}dt-\frac{1}{2}\sigma^2V_tdt.
\end{eqnarray*}
Note $V$ appears only in the $dt$ term, so multiply the integration
factor $e^{\frac{1}{2}\sigma^2t}$ on both sides of the equation, we
get
\[
d(e^{\frac{1}{2}\sigma^2 t}V_t)=ae^{-rt-\sigma \widetilde
W_t+\frac{1}{2}\sigma^2 t}dt.
\]Set $Y_t=e^{\sigma\widetilde W_t+(r-\frac{1}{2}\sigma^2)t}$, we
have $d(S_t/Y_t)=adt/Y_t$. So
$S_t=Y_t(S_0+\int_0^t\frac{ads}{Y_s}).$
\end{remark}


(iii)\begin{proof} \begin{eqnarray*} \widetilde E[S_T|{\cal F}_t]
&=&S_0\widetilde E[Y_T|{\cal F}_t]+\widetilde
E\left[Y_T\int_0^t\frac{a}{Y_s}ds+Y_T\int_t^T\frac{a}{Y_s}ds|{\cal F}_t\right]\\
&=&S_0\widetilde E[Y_T|{\cal F}_t]+\int_0^t\frac{a}{Y_s}ds\widetilde
E[Y_T|{\cal F}_t]+a\int_t^T\widetilde E\left[\frac{Y_T}{Y_s}|{\cal
F}_t\right]ds\\
&=&S_0Y_t\widetilde E[Y_{T-t}]+\int_0^t\frac{a}{Y_s}dsY_t\widetilde
E[Y_{T-t}]+a\int_t^T\widetilde E[Y_{T-s}]ds\\
&=&\left(S_0+\int_0^t\frac{a}{Y_s}ds\right)Y_te^{r(T-t)}+a\int_t^Te^{r(T-s)}ds\\
&=&\left(S_0+\int_0^t\frac{ads}{Y_s}\right)Y_te^{r(T-t)}-\frac{a}{r}(1-e^{r(T-t)}).
\end{eqnarray*}
In particular, $\widetilde E[S_T]=S_0e^{rT}-\frac{a}{r}(1-e^{rT})$.
\end{proof}

(iv)\begin{proof}
\begin{eqnarray*}
d\widetilde E[S_T|{\cal
F}_t]&=&ae^{r(T-t)}dt+\left(S_0+\int_0^t\frac{ads}{Y_s}\right)(e^{r(T-t)}dY_t-rY_te^{r(T-t)}dt)+\frac{a}{r}e^{r(T-t)}(-r)dt\\
&=&\left(S_0+\int_0^t\frac{ads}{Y_s}\right)e^{r(T-t)}\sigma
Y_td\widetilde W_t.
\end{eqnarray*}
So $\widetilde E[S_T|{\cal F}_t]$ is a $\widetilde P$-martingale. As
we have argued at the beginning of the solution, risk-neutral
pricing is valid even in the presence of cost of carry. So by an
argument similar to that of \S 5.6.2, the process $\widetilde
E[S_T|{\cal F}_t]$ is the futures price process for the commodity.
\end{proof}

(v)\begin{proof} We solve the equation $\widetilde
E[e^{-r(T-t)}(S_T-K)|{\cal F}_t]=0$ for $K$, and get $K=\widetilde E
[S_T|{\cal F}_t]$. So $For_S(t,T)=Fut_S(t,T)$.
\end{proof}

(vi)\begin{proof}We follow the hint. First, we solve the SDE
\[
\begin{cases}
dX_t=dS_t-adt+r(X_t-S_t)dt\\
X_0=0.
\end{cases}
\]By our analysis in part (i),
$d(e^{-rt}X_t)=d(e^{-rt}S_t)-ae^{-rt}dt$. Integrate from $0$ to $t$
on both sides, we get
$X_t=S_t-S_0e^{rt}+\frac{a}{r}(1-e^{rt})=S_t-S_0e^{rt}-\frac{a}{r}(e^{rt}-1)$.
In particular, $X_T=S_T-S_0e^{rT}-\frac{a}{r}(e^{rT}-1)$. Meanwhile,
$For_S(t,T)=Fut_s(t,T)=\widetilde E[S_T|{\cal
F}_t]=\left(S_0+\int_0^t\frac{ads}{Y_s}\right)Y_te^{r(T-t)}-\frac{a}{r}(1-e^{r(T-t)})$.
So $For_S(0,T)=S_0e^{rT}-\frac{a}{r}(1-e^{rT})$ and hence
$X_T=S_T-For_S(0,T)$. After the agent delivers the commodity, whose
value is $S_T$, and receives the forward price $For_S(0,T)$, the
portfolio has exactly zero value.
\end{proof}

\medskip

\section{Connections with Partial Differential Equations}

$\bigstar$ {\bf Comments}:

\medskip

1) A rigorous presentation of (strong) Markov property can be found in Revuz and Yor \cite{RY98}, Chapter III.

\medskip

2) A rigorous presentation of the Feymann-Kac formula can be found in \O ksendal \cite[page~143]{Oksendal03} (also see 钱敏平等 \cite[page~236]{QG97}). To reconcile the version presented in this book and the one in \O ksendal \cite{Oksendal03} (Theorem 8.2.1), we note for the undiscounted version, in this book
\[
g(t,X(t)) = {\mathbb E}[h(X(T))|{\cal F}(t)] = {\mathbb E}^{X(t)}[h(X(T-t))]
\]
while in \O ksendal \cite{Oksendal03}
\[
v(t,x)= E^x[h(X_t)].
\]
So $v(t,x)=g(T-t,x)$. The discounted version can connected similarly.

\medskip

3) {\it Hedging equation}. Recall the SDE satisfied by the value process $X(t)$ of a self-financing portfolio is (see formula (5.4.21) and (5.4.22) on page 230)
\begin{eqnarray*}
d[D(t)X(t)]
&=& D(t) [dX(t)-R(t)X(t)dt] \\
&=& D(t) \left\{ \sum_{i=1}^m \Delta_i(t) dS_i(t) + R(t) \left[ X(t)-\sum_{i=1}^m \Delta_i(t)S_i(t)\right]dt -R(t)X(t)dt \right\} \\
&=& D(t) \left[ \sum_{i=1}^m \Delta_i(t) dS_i(t) - R(t) \sum_{i=1}^m \Delta_i(t)S_i(t) dt  \right] \\
&=& \sum_{i=1}^m \Delta_i(t) d[D(t)S_i(t)].
\end{eqnarray*}
Under the multidimensional market model (formula (5.4.6) on page 226)
\[
dS_i(t) = \alpha_i(t) S_i(t)dt + S_i(t) \sum_{j=1}^d \sigma_{ij}(t) dW_j(t), \; i = 1, \cdots, m.
\]
and assuming the existence of the risk-neutral measure (that is, the {\it market price of risk equations} has a solution (formula (5.4.18) on page 228): $\alpha_i(t) - R(t) = \sum_{j=1}^d \sigma_{ij}(t) \Theta_j(t), \; i= 1, \cdots, m$), $D(t)S_i(t)$ is a martingale under the risk-neutral measure (formula (5.4.17) on page 228) satisfying the SDE
\[
d[D(t)S_i(t)] = D(t) S_i(t) \sum_{j=1}^d \sigma_{ij}(t) d\widetilde W_j(t), \; dS_i(t) = R(t)S_i(t)dt + S_i(t) \sum_{j=1}^d \sigma_{ij}(t)d\widetilde W_j(t), \; i=1, \cdots, m.
\]
Consequently, $d[D(t)X(t)]= \sum_{i=1}^m \Delta_i(t) d[D(t)S_i(t)]$ becomes
\begin{eqnarray*}
d[D(t)X(t)] = D(t)\sum_{i=1}^m \left[\Delta_i(t)S_i(t) \sum_{j=1}^d \sigma_{ij}(t) d\widetilde W_j(t)\right].
\end{eqnarray*}
If we further assume $X(t)$ has the form $v(t,S(t))$, then the above equation becomes
\begin{eqnarray*}
& & D(t) \left[v_t(t,S(t))dt+\sum_{i=1}^m v_{x_i}(t,S(t))dS_i(t) + \frac{1}{2}\sum_{k,l=1}^m v_{x_kx_l}(t,S(t))dS_k(t)dS_l(t)\right] -R(t)v(t,X(t)) dt \\
&=& (\cdots) dt + D(t) \sum_{i=1}^m v_{x_i}(t,S(t))\left[S_i(t) \sum_{j=1}^d \sigma_{ij}(t)d\widetilde W_j(t)\right] \\
&=& D(t)\sum_{i=1}^m \left[\Delta_i(t)S_i(t) \sum_{j=1}^d \sigma_{ij}(t) d\widetilde W_j(t)\right].
\end{eqnarray*}
Equating the coefficient of each $d\widetilde W_j(t)$, we have the {\it hedging equation}
\[
\sum_{i=1}^m v_{x_i}(t,S(t))S_i(t)\sigma_{ij}(t) = \sum_{i=1}^m \Delta_i(t)S_i(t)\sigma_{ij}(t), \; j=1, \cdots, d.
\]
{\it One} solution of the hedging equation is
\[
\Delta_i(t) = v_{x_i}(t,S(t)), \; i=1, \cdots, m.
\]
\bigskip

\noindent $\blacktriangleright$ {\bf Exercise 6.1.} Consider the stochastic differential equation
\[
dX(u) =
\]

\smallskip

(i) \begin{proof} $Z_t=1$ is obvious. Note the form
of $Z$ is similar to that of a geometric Brownian motion. So by
It\^{o}'s formula, it is easy to obtain
$dZ_u=b_uZ_udu+\sigma_uZ_udW_u$, $u\ge t$.
\end{proof}

(ii)\begin{proof} If $X_u=Y_uZ_u$ ($u\ge t$), then
$X_t=Y_tZ_t=x\cdot 1=x$ and
\begin{eqnarray*}
dX_u&=&Y_udZ_u+Z_udY_u+dY_uZ_u\\
&=&Y_u(b_uZ_udu+\sigma_uZ_udW_u)+Z_u\left(\frac{a_u-\sigma_u\gamma_u}{Z_u}du+\frac{\gamma_u}{Z_u}dW_u\right)+\sigma_uZ_u\frac{\gamma_u}{Z_u}du\\
&=&[Y_ub_uZ_u+(a_u-\sigma_u\gamma_u)+\sigma_u\gamma_u]du+(\sigma_uZ_uY_u+\gamma_u)dW_u\\
&=&(b_uX_u+a_u)du+(\sigma_uX_u+\gamma_u)dW_u.
\end{eqnarray*}
\end{proof}

\begin{remark} To see how to find the above solution, we manipulate
the equation (6.2.4) as follows. First, to remove the term
$b_uX_udu$, we multiply on both sides of (6.2.4) the integrating
factor $e^{-\int_t^ub_vdv}$. Then
\[
d(X_ue^{-\int_t^ub_vdv})=e^{-\int_t^ub_vdv}(a_udu+(\gamma_u+\sigma_uX_u)dW_u).
\]
Let $\bar X_u=e^{-\int_t^ub_vdv}X_u$, $\bar
a_u=e^{-\int_t^ub_vdv}a_u$ and $\bar
\gamma_u=e^{-\int_t^ub_vdv}\gamma_u$, then $\bar X$ satisfies the
SDE
\[
d\bar X_u=\bar a_udu+(\bar \gamma_u+\sigma_u\bar X_u)dW_u=(\bar
a_udu+\bar \gamma_udW_u)+\sigma_u\bar X_udW_u.
\]
To deal with the term $\sigma_u\bar X_udW_u$, we consider $\hat
X_u=\bar X_ue^{-\int_t^u\sigma_vdW_v}$. Then
\begin{eqnarray*}
d\hat X_u &=&e^{-\int_t^u\sigma_vdW_v}[(\bar a_udu+\bar
\gamma_udW_u)+\sigma_u\bar X_udW_u]+\bar
X_u\left(e^{-\int_t^u\sigma_vdW_v}(-\sigma_u)dW_u+\frac{1}{2}e^{-\int_t^u\sigma_vdW_v}\sigma_u^2du\right)\\
& &+(\bar\gamma_u+\sigma_u\bar
X_u)(-\sigma_u)e^{-\int_t^u\sigma_vdW_v}du\\
&=&\hat a_udu+\hat \gamma_udW_u+\sigma_u\hat X_udW_u-\sigma_u\hat
X_udW_u+\frac{1}{2}\hat
X_u\sigma_u^2du-\sigma_u(\hat\gamma_u+\sigma_u\hat X_u)du\\
&=&(\hat a_u-\sigma_u\hat \gamma_u-\frac{1}{2}\hat
X_u\sigma_u^2)du+\hat \gamma_udW_u,
\end{eqnarray*}where $\hat a_u=\bar a_u e^{-\int_t^u\sigma_vdW_v}$
and $\hat \gamma_u=\bar\gamma_ue^{-\int_t^u\sigma_vdW_v}$. Finally,
use the integrating factor $e^{\int_t^u\frac{1}{2}\sigma_v^2dv}$, we
have
\[
d\left(\hat
X_ue^{\frac{1}{2}\int_t^u\sigma_v^2dv}\right)=e^{\frac{1}{2}\int_t^u\sigma_v^2dv}(d\hat
X_u+\hat X_u\cdot
\frac{1}{2}\sigma_u^2du)=e^{\frac{1}{2}\int_t^u\sigma_v^2dv}[(\hat
a_u-\sigma_u\hat \gamma_u)du+\hat \gamma_udW_u].
\]
Write everything back into the original $X$, $a$ and $\gamma$, we
get
\[
d\left(X_ue^{-\int_t^ub_vdv-\int_t^u\sigma_vdW_v+\frac{1}{2}\int_t^u\sigma^2_vdv}\right)=e^{\frac{1}{2}\int_t^u\sigma_v^2dv-\int_t^u\sigma_vdW_v-\int_t^ub_vdv}[(a_u-\sigma_u\gamma_u)du+\gamma_udW_u],
\]i.e.\[
d\left(\frac{X_u}{Z_u}\right)=\frac{1}{Z_u}[(a_u-\sigma_u\gamma_u)du+\gamma_udW_u]=dY_u.
\]This inspired us to try $X_u=Y_uZ_u$.
\end{remark}


\noindent 6.2. (i)\begin{proof} The portfolio is self-financing, so
for any $t\le T_1$, we have
\[
dX_t=\Delta_1(t)df(t,R_t,T_1)+\Delta_2(t)df(t,R_t,T_2)+R_t(X_t-\Delta_1(t)f(t,R_t,T_1)-\Delta_2(t)f(t,R_t,T_2))dt,
\]and
\begin{eqnarray*}
& &d(D_tX_t)\\
 &=&-R_tD_tX_tdt+D_tdX_t\\
&=&D_t[\Delta_1(t)df(t,R_t,T_1)+\Delta_2(t)df(t,R_t,T_2)-R_t(\Delta_1(t)f(t,R_t,T_1)+\Delta_2(t)f(t,R_t,T_2))dt]\\
&=&D_t[\Delta_1(t)\left(f_t(t,R_t,T_1)dt+f_r(t,R_t,T_1)dR_t+\frac{1}{2}f_{rr}(t,R_t,T_1)\gamma^2(t,R_t)dt\right)\\
&
&+\Delta_2(t)\left(f_t(t,R_t,T_2)dt+f_r(t,R_t,T_2)dR_t+\frac{1}{2}f_{rr}(t,R_t,T_2)\gamma^2(t,R_t)dt\right)\\
& &-R_t(\Delta_1(t)f(t,R_t,T_1)+\Delta_2(t)f(t,R_t,T_2))dt]\\
&=&\Delta_1(t)D_t[-R_tf(t,R_t,T_1)+f_t(t,R_t,T_1)+\alpha(t,R_t)f_r(t,R_t,T_1)+\frac{1}{2}\gamma^2(t,R_t)f_{rr}(t,R_t,T_1)]dt\\
& &+\Delta_2(t)D_t[-R_tf(t,R_t,T_2)+f_t(t,R_t,T_2)+\alpha(t,R_t)f_r(t,R_t,T_2)+\frac{1}{2}\gamma^2(t,R_t)f_{rr}(t,R_t,T_2)]dt\\
& &+D_t\gamma(t,R_t)[D_t\gamma(t,R_t)[\Delta_1(t)f_r(t,R_t,T_1)+\Delta_2(t)f_r(t,R_t,T_2)]]dW_t\\
&=&\Delta_1(t)D_t[\alpha(t,R_t)-\beta(t,R_t,T_1)]f_r(t,R_t,T_1)dt+\Delta_2(t)D_t[\alpha(t,R_t)-\beta(t,R_t,T_2)]f_r(t,R_t,T_2)dt\\
&
&+D_t\gamma(t,R_t)[\Delta_1(t)f_r(t,R_t,T_1)+\Delta_2(t)f_r(t,R_t,T_2)]dW_t.
\end{eqnarray*}
\end{proof}

(ii)\begin{proof} Let $\Delta_1(t)=S_tf_r(t,R_t,T_2)$ and
$\Delta_2(t)=-S_tf_r(t,R_t,T_1)$, then
\begin{eqnarray*}
d(D_tX_t)&=&D_tS_t[\beta(t,R_t,T_2)-\beta(t,R_t,T_1)]f_r(t,R_t,T_1)f_r(t,R_t,T_2)dt\\
&=&D_t|[\beta(t,R_t,T_1)-\beta(t,R_t,T_2)]f_r(t,R_t,T_1)f_r(t,R_t,T_2)|dt.
\end{eqnarray*}
Integrate from $0$ to $T$ on both sides of the above equation, we
get
\[
D_TX_T-D_0X_0=\int_0^TD_t|[\beta(t,R_t,T_1)-\beta(t,R_t,T_2)]f_r(t,R_t,T_1)f_r(t,R_t,T_2)|dt.
\]
If $\beta(t,R_t,T_1)\ne \beta(t,R_t,T_2)$ for some $t\in[0,T]$,
under the assumption that $f_r(t,r,T)\ne 0$ for all values of $r$
and $0\le t\le T$, $D_TX_T-D_0X_0>0$. To avoid arbitrage (see, for
example, Exercise 5.7), we must have for a.s. $\omega$,
$\beta(t,R_t,T_1)=\beta(t,R_t,T_2)$, $\forall t\in [0,T]$. This
implies $\beta(t,r,T)$ does not depend on $T$.
\end{proof}

(iii) \begin{proof} In (6.9.4), let $\Delta_1(t)=\Delta(t)$, $T_1=T$
and $\Delta_2(t)=0$, we get
\begin{eqnarray*}
d(D_tX_t)&=&\Delta(t)D_t\left[-R_tf(t,R_t,T)+f_t(t,R_t,T)+\alpha(t,R_t)f_r(t,R_t,T)+\frac{1}{2}\gamma^2(t,R_t)f_{rr}(t,R_t,T)\right]dt\\
& &+D_t\gamma(t,R_t)\Delta(t)f_r(t,R_t,T)dW_t.
\end{eqnarray*}
This is formula (6.9.5).

If $f_r(t,r,T)=0$, then
$d(D_tX_t)=\Delta(t)D_t\left[-R_tf(t,R_t,T)+f_t(t,R_t,T)+\frac{1}{2}\gamma^2(t,R_t)f_{rr}(t,R_t,T)\right]dt$.
We choose
$\Delta(t)=\mbox{sign}\left\{\left[-R_tf(t,R_t,T)+f_t(t,R_t,T)+\frac{1}{2}\gamma^2(t,R_t)f_{rr}(t,R_t,T)\right]\right\}.$
To avoid arbitrage in this case, we must have
$f_t(t,R_t,T)+\frac{1}{2}\gamma^2(t,R_t)f_{rr}(t,R_t,T)=R_tf(t,R_t,T)$,
or equivalently, for any $r$ in the range of $R_t$,
$f_t(t,r,T)+\frac{1}{2}\gamma^2(t,r)f_{rr}(t,r,T)=rf(t,r,T)$.
\end{proof}

\noindent 6.3. \begin{proof} We note
\[
\frac{d}{ds}\left[e^{-\int_0^sb_vdv}C(s,T)\right]=e^{-\int_0^sb_vdv}[C(s,T)(-b_s)+b_sC(s,T)-1]=-e^{-\int_0^sb_vdv}.
\]So integrate on both sides of the equation from t to T, we obtain
\[
e^{-\int_0^Tb_vdv}C(T,T)-e^{-\int_0^tb_vdv}C(t,T)=-\int_t^Te^{-\int_0^sb_vdv}ds.
\]Since $C(T,T)=0$, we have
$C(t,T)=e^{\int_0^tb_vdv}\int_t^Te^{-\int_0^sb_vdv}ds=\int_t^Te^{\int_s^tb_vdv}ds$.
Finally, by $A'(s,T)=-a(s)C(s,T)+\frac{1}{2}\sigma^2(s)C^2(s,T)$, we
get
\[
A(T,T)-A(t,T)=-\int_t^Ta(s)C(s,T)ds+\frac{1}{2}\int_t^T\sigma^2(s)C^2(s,T)ds.
\]Since $A(T,T)=0$, we have
$A(t,T)=\int_t^T(a(s)C(s,T)-\frac{1}{2}\sigma^2(s)C^2(s,T))ds$.
\end{proof}

\noindent 6.4. (i)\begin{proof} By the definition of $\varphi$, we
have
\[
\varphi'(t)=e^{\frac{1}{2}\sigma^2\int_t^TC(u,T)du}\frac{1}{2}\sigma^2(-1)C(t,T)=-\frac{1}{2}\varphi(t)\sigma^2C(t,T).
\]
So $C(t,T)=-\frac{2\varphi'(t)}{\phi(t)\sigma^2}$. Differentiate
both sides of the equation
$\varphi'(t)=-\frac{1}{2}\varphi(t)\sigma^2C(t,T)$, we get
\begin{eqnarray*}
\varphi''(t)&=&-\frac{1}{2}\sigma^2[\varphi'(t)C(t,T)+\varphi(t)C'(t,T)]\\
&=&-\frac{1}{2}\sigma^2[-\frac{1}{2}\varphi(t)\sigma^2C^2(t,T)+\varphi(t)C'(t,T)]\\
&=&\frac{1}{4}\sigma^4\varphi(t)C^2(t,T)-\frac{1}{2}\sigma^2\varphi(t)C'(t,T).
\end{eqnarray*}
So
$C'(t,T)=\left[\frac{1}{4}\sigma^4\varphi(t)C^2(t,T)-\varphi''(t)\right]/\frac{1}{2}\varphi(t)\sigma^2=\frac{1}{2}\sigma^2C^2(t,T)-\frac{2\varphi''(t)}{\sigma^2\varphi(t)}$.
\end{proof}

(ii)\begin{proof}Plug formulas (6.9.8) and (6.9.9) into (6.5.14), we
get
\[
-\frac{2\varphi''(t)}{\sigma^2\varphi(t)}+\frac{1}{2}\sigma^2C^2(t,T)=b(-1)\frac{2\varphi'(t)}{\sigma^2\varphi(t)}+\frac{1}{2}\sigma^2C^2(t,T)-1,
\]i.e. $\varphi''(t)-b\varphi'(t)-\frac{1}{2}\sigma^2\varphi(t)=0$.
\end{proof}

(iii)\begin{proof} The characteristic equation of
$\varphi''(t)-b\varphi'(t)-\frac{1}{2}\sigma^2\varphi(t)=0$ is
$\lambda^2-b\lambda-\frac{1}{2}\sigma^2=0$, which gives two roots
$\frac{1}{2}(b\pm\sqrt{b^2+2\sigma^2})=\frac{1}{2}b\pm \gamma$ with
$\gamma=\frac{1}{2}\sqrt{b^2+2\sigma^2}$. Therefore by standard
theory of ordinary differential equations, a general solution of
$\varphi$ is $\varphi(t)=e^{\frac{1}{2}bt}(a_1e^{\gamma
t}+a_2e^{-\gamma t})$ for some constants $a_1$ and $a_2$. It is then
easy to see that we can choose appropriate constants $c_1$ and $c_2$
so that
\[
\varphi(t)=\frac{c_1}{\frac{1}{2}b+\gamma}e^{-(\frac{1}{2}b+\gamma)(T-t)}-\frac{c_2}{\frac{1}{2}b-\gamma}e^{-(\frac{1}{2}b-\gamma)(T-t)}.
\]
\end{proof}

(iv) \begin{proof} From part (iii), it is easy to see
$\varphi'(t)=c_1e^{-(\frac{1}{2}b+\gamma)(T-t)}-c_2e^{-(\frac{1}{2}b-\gamma)(T-t)}$.
In particular, \[
0=C(T,T)=-\frac{2\varphi'(T)}{\sigma^2\varphi(T)}=-\frac{2(c_1-c_2)}{\sigma^2\varphi(T)}.
\]
So $c_1=c_2$.
\end{proof}

(v)\begin{proof} We first recall the definitions and properties of
$\sinh$ and $\cosh$:
\[
\sinh z=\frac{e^z-e^{-z}}{2},\;\cosh z=\frac{e^z+e^{-z}}{2},\;
(\sinh z)'=\cosh z,\;\mbox{and}\; (\cosh z)'=\sinh z.
\]Therefore
\begin{eqnarray*}
\varphi(t) &=&
c_1e^{-\frac{1}{2}b(T-t)}\left[\frac{e^{-\gamma(T-t)}}{\frac{1}{2}b+\gamma}-\frac{e^{\gamma(T-t)}}{\frac{1}{2}b-\gamma}\right]\\
&=&c_1e^{-\frac{1}{2}b(T-t)}\left[\frac{\frac{1}{2}b-\gamma}{\frac{1}{4}b^2-\gamma^2}e^{-\gamma(T-t)}-\frac{\frac{1}{2}b+\gamma}{\frac{1}{4}b^2-\gamma^2}e^{\gamma(T-t)}\right]\\
&=&\frac{2c_1}{\sigma^2}e^{-\frac{1}{2}b(T-t)}\left[-(\frac{1}{2}b-\gamma)e^{-\gamma(T-t)}+(\frac{1}{2}b+\gamma)e^{\gamma(T-t)}\right]\\
&=&\frac{2c_1}{\sigma^2}e^{-\frac{1}{2}b(T-t)}[b\sinh(\gamma(T-t))+2\gamma\cosh(\gamma(T-t))].
\end{eqnarray*}
and
\begin{eqnarray*}
\varphi'(t)&=&\frac{1}{2}b\cdot\frac{2c_1}{\sigma^2}e^{-\frac{1}{2}b(T-t)}[b\sinh(\gamma(T-t))+2\gamma\cosh(\gamma(T-t))]\\
& &+\frac{2c_1}{\sigma^2}e^{-\frac{1}{2}b(T-t)}[-\gamma
b\cosh (\gamma(T-t))-2\gamma^2\sinh (\gamma(T-t))]\\
&=&2c_1e^{-\frac{1}{2}b(T-t)}\left[\frac{b^2}{2\sigma^2}\sinh
(\gamma(T-t))+\frac{b\gamma}{\sigma^2}\cosh(\gamma(T-t))-\frac{b\gamma}{\sigma^2}\cosh(\gamma(T-t)) \right.\\
& & \left.-\frac{2\gamma^2}{\sigma^2}\sinh(\gamma(T-t))\right]\\
&=&2c_1e^{-\frac{1}{2}b(T-t)}\frac{b^2-4\gamma^2}{2\sigma^2}\sinh(\gamma(T-t))\\
&=&-2c_1e^{-\frac{1}{2}b(T-t)}\sinh(\gamma(T-t)).
\end{eqnarray*}
This implies
\[
C(t,T)=-\frac{2\varphi'(t)}{\sigma^2\varphi(t)}=\frac{\sinh(\gamma(T-t))}{\gamma\cosh(\gamma(T-t))+\frac{1}{2}b\sinh(\gamma(T-t))}.
\]
\end{proof}

(vi) \begin{proof} By (6.5.15) and (6.9.8),
$A'(t,T)=\frac{2a\varphi'(t)}{\sigma^2\varphi(t)}$. Hence
\[
A(T,T)-A(t,T)=\int_t^T\frac{2a\varphi'(s)}{\sigma^2\varphi(s)}ds=\frac{2a}{\sigma^2}\ln\frac{\varphi(T)}{\varphi(t)},
\]and
\[
A(t,T)=-\frac{2a}{\sigma^2}\ln\frac{\varphi(T)}{\varphi(t)}=-\frac{2a}{\sigma^2}\ln\left[\frac{\gamma
e^{\frac{1}{2}b(T-t)}}{\gamma\cosh(\gamma(T-t))+\frac{1}{2}b\sinh(\gamma(T-t))}\right].
\]
\end{proof}

\noindent 6.5. (i)\begin{proof} Since
$g(t,X_1(t),X_2(t))=E[h(X_1(T),X_2(T))|{\cal F}_t]$ and
\[
e^{-rt}f(t,X_1(t),X_2(t))=E[e^{-rT}h(X_1(T),X_2(T))|{\cal F}_t],
\]
iterated conditioning argument shows $g(t,X_1(t),X_2(t))$ and
$e^{-rt}f(t,X_1(t),X_2(t))$ ar both martingales.
\end{proof}

(ii) and (iii)\begin{proof} We note
\begin{eqnarray*}
& &dg(t,X_1(t),X_2(t))\\
&=&g_tdt+g_{x_1}dX_1(t)+g_{x_2}dX_2(t)+\frac{1}{2}g_{x_1x_2}dX_1(t)dX_1(t)+\frac{1}{2}g_{x_2x_2}dX_2(t)dX_2(t)+g_{x_1x_2}dX_1(t)dX_2(t)\\
&=&\left[g_t+g_{x_1}\beta_1+g_{x_2}\beta_2+\frac{1}{2}g_{x_1x_1}(\gamma_{11}^2+\gamma^2_{12}+2\rho\gamma_{11}\gamma_{12})+g_{x_1x_2}(\gamma_{11}\gamma_{21}+\rho\gamma_{11}\gamma_{22}+\rho\gamma_{12}\gamma_{21}
+\gamma_{12}\gamma_{22})\right.\\
& &\left.+\frac{1}{2}
g_{x_2x_2}(\gamma^2_{21}+\gamma_{22}^2+2\rho\gamma_{21}\gamma_{22})\right]dt+\mbox{martingale
part}.
\end{eqnarray*}
So we must have
\begin{eqnarray*}
&
&g_t+g_{x_1}\beta_1+g_{x_2}\beta_2+\frac{1}{2}g_{x_1x_1}(\gamma_{11}^2+\gamma^2_{12}+2\rho\gamma_{11}\gamma_{12})+g_{x_1x_2}(\gamma_{11}\gamma_{21}+\rho\gamma_{11}\gamma_{22}+\rho\gamma_{12}\gamma_{21}
+\gamma_{12}\gamma_{22})\\
& &+\frac{1}{2}
g_{x_2x_2}(\gamma^2_{21}+\gamma_{22}^2+2\rho\gamma_{21}\gamma_{22})=0.
\end{eqnarray*}
Taking $\rho=0$ will give part (ii) as a special case. The PDE for
$f$ can be similarly obtained.
\end{proof}

\noindent 6.6. (i)\begin{proof} Multiply $e^{\frac{1}{2}bt}$ on both
sides of (6.9.15), we get
\[
d(e^{\frac{1}{2}bt}X_j(t))=e^{\frac{1}{2}bt}\left(X_j(t)\frac{1}{2}bdt+(-\frac{b}{2}X_j(t)dt+\frac{1}{2}\sigma
dW_j(t)\right)=e^{\frac{1}{2}bt}\frac{1}{2}\sigma dW_j(t).
\]So
$e^{\frac{1}{2}bt}X_j(t)-X_j(0)=\frac{1}{2}\sigma\int_0^te^{\frac{1}{2}bu}dW_j(u)$
and
$X_j(t)=e^{-\frac{1}{2}bt}\left(X_j(0)+\frac{1}{2}\sigma\int_0^te^{\frac{1}{2}bu}dW_j(u)\right)$.
By Theorem 4.4.9, $X_j(t)$ is normally distributed with mean
$X_j(0)e^{-\frac{1}{2}bt}$ and variance
$\frac{e^{-bt}}{4}\sigma^2\int_0^te^{bu}du=\frac{\sigma^2}{4b}(1-e^{-bt})$.
\end{proof}

(ii)\begin{proof} Suppose $R(t)=\sum_{j=1}^dX_j^2(t)$, then
\begin{eqnarray*}
dR(t)&=&\sum_{j=1}^d(2X_j(t)dX_j(t)+dX_j(t)dX_j(t))\\
&=&\sum_{j=1}^d\left(2X_j(t)dX_j(t)+\frac{1}{4}\sigma^2dt\right)\\
&=&\sum_{j=1}^d\left(-bX_j^2(t)dt+\sigma
X_j(t)dW_j(t)+\frac{1}{4}\sigma^2dt\right)\\
&=&\left(\frac{d}{4}\sigma^2-bR(t)\right)dt+\sigma\sqrt{R(t)}\sum_{j=1}^d\frac{X_j(t)}{\sqrt{R(t)}}dW_j(t).
\end{eqnarray*}
Let $B(t)=\sum_{j=1}^d\int_0^t\frac{X_j(s)}{\sqrt{R(s)}}dW_j(s)$,
then $B$ is a local martingale with
$dB(t)dB(t)=\sum_{j=1}^d\frac{X_j^2(t)}{R(t)}dt=dt$. So by
L\'{e}vy's Theorem, $B$ is a Brownian motion. Therefore
$dR(t)=(a-bR(t))dt+\sigma\sqrt{R(t)}dB(t)$
$(a:=\frac{d}{4}\sigma^2)$ and $R$ is a CIR interest rate process.
\end{proof}

(iii)\begin{proof} By (6.9.16), $X_j(t)$ is dependent on $W_j$ only
and is normally distributed with mean $e^{-\frac{1}{2}bt}X_j(0)$ and
variance $\frac{\sigma^2}{4b}[1-e^{-bt}]$. So $X_1(t)$, $\cdots$,
$X_d(t)$ are i.i.d. normal with the same mean $\mu(t)$ and variance
$v(t)$.
\end{proof}

(iv)\begin{proof}
\begin{eqnarray*}
E\left[e^{uX_j^2(t)}\right]&=&\int_{-\infty}^{\infty}e^{ux^2}\frac{e^{-\frac{(x-\mu(t))^2}{2v(t)}}dx}{\sqrt{2\pi
v(t)}}\\
&=&\int_{-\infty}^{\infty}\frac{e^{-\frac{(1-2uv(t))x^2-2\mu(t)x+\mu^2(t)}{2v(t)}}}{\sqrt{2\pi
v(t)}}dx\\
&=&\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi v(t)}}e^{-\frac{\left(x-\frac{\mu(t)}{1-2uv(t)}\right)^2+\frac{\mu^2(t)}{1-2uv(t)}-\frac{\mu^2(t)}{(1-2uv(t))^2}}{2v(t)/(1-2uv(t))}}dx\\
&=&\int_{-\infty}^{\infty}\frac{\sqrt{1-2uv(t)}}{\sqrt{2\pi
v(t)}}e^{-\frac{\left(x-\frac{\mu(t)}{1-2uv(t)}\right)^2}{2v(t)/(1-2uv(t))}}dx
\cdot
\frac{e^{-\frac{\mu^2(t)(1-2uv(t))-\mu^2(t)}{2v(t)(1-2uv(t))}}}{\sqrt{1-2uv(t)}}\\
&=&\frac{e^{-\frac{u\mu^2(t)}{1-2uv(t)}}}{\sqrt{1-2uv(t)}}.
\end{eqnarray*}
\end{proof}


(v)\begin{proof} By $R(t)=\sum_{j=1}^dX_j^2(t)$ and the fact
$X_1(t)$, $\cdots$, $X_d(t)$ are i.i.d.,
\[
E[e^{uR(t)}]=(E[e^{uX_1^2(t)}])^d=(1-2uv(t))^{-\frac{d}{2}}e^{\frac{ud\mu^2(t)}{1-2uv(t)}}=(1-2uv(t))^{-\frac{2a}{\sigma^2}}e^{-\frac{e^{-bt}uR(0)}{1-2uv(t)}}.
\]
\end{proof}

\noindent 6.7. (i)\begin{proof} $e^{-rt}c(t,S_t,V_t)=\widetilde
E[e^{-rT}(S_T-K)^+|{\cal F}_t]$ is a martingale by iterated
conditioning argument. Since
\begin{eqnarray*}
&
&d(e^{-rt}c(t,S_t,V_t))\\
&=&e^{-rt}\left[c(t,S_t,V_t)(-r)+c_t(t,S_t,V_t)+c_s(t,S_t,V_t)rS_t+c_v(t,S_t,V_t)(a-bV_t)+\frac{1}{2}c_{ss}(t,S_t,V_t)V_tS_t^2+\right.\\
& & \left.\frac{1}{2}
c_{vv}(t,S_t,V_t)\sigma^2V_t+c_{sv}(t,S_t,V_t)\sigma
V_tS_t\rho\right]dt+\mbox{martingale part},
\end{eqnarray*}
we conclude
$rc=c_t+rsc_s+c_v(a-bv)+\frac{1}{2}c_{ss}vs^2+\frac{1}{2}c_{vv}\sigma^2v+c_{sv}\sigma
sv\rho$. This is equation (6.9.26).
\end{proof}

(ii)\begin{proof}Suppose $c(t,s,v)=sf(t,\log
s,v)-e^{-r(T-t)}Kg(t,\log s, v)$, then
\begin{eqnarray*}
& &c_t=sf_t(t,\log s,v)-re^{-r(T-t)}Kg(t,\log
s,v)-e^{-r(T-t)}Kg_t(t,\log s,v),\\
& &c_s=f(t,\log s,v)+sf_s(t,\log
s,v)\frac{1}{s}-e^{-r(T-t)}Kg_s(t,\log
s,v)\frac{1}{s},\\
& &c_v=sf_v(t,\log s, v)-e^{-r(T-t)}Kg_v(t,\log s,v),\\
& &c_{ss}=f_s(t,\log s,v)\frac{1}{s}+f_{ss}(t,\log s,
v)\frac{1}{s}-e^{-r(T-t)}Kg_{ss}(t,\log
s,v)\frac{1}{s^2}+e^{-r(T-t)}Kg_s(t,\log s, v)\frac{1}{s^2},\\
& &c_{sv}=f_v(t,\log s,v)+f_{sv}(t,\log
s,v)-e^{-r(T-t)}\frac{K}{s}g_{sv}(t,\log s,v),\\
& &c_{vv}=sf_{vv}(t,\log s,v)-e^{-r(T-t)}Kg_{vv}(t,\log s,v).
\end{eqnarray*}
So
\begin{eqnarray*}
& &c_t+rsc_s+(a-b v )c_v+\frac{1}{2}s^2vc_{ss}+\rho\sigma
svc_{sv}+\frac{1}{2}\sigma^2vc_{vv}\\
&=&sf_t-re^{-r(T-t)}Kg-e^{-r(T-t)}Kg_t+rsf+rsf_s-rKe^{-r(T-t)}g_s+(a-bv)(sf_v-e^{-r(T-t)}Kg_v)\\
& &+\frac{1}{2}s^2v\left[-\frac{1}{s}f_s+\frac{1}{s}f_{ss}
-e^{-r(T-t)}\frac{K}{s^2}g_{ss}+e^{-r(T-t)}K\frac{g_s}{s^2}\right]+\rho\sigma
sv\left(f_v+f_{sv}-e^{-r(T-t)}\frac{K}{s}g_{sv}\right)\\
& &+\frac{1}{2}\sigma^2v(sf_{vv}-e^{-r(T-t)}Kg_{vv})\\
&=&s\left[f_t+(r+\frac{1}{2}v)f_s+(a-bv+\rho\sigma
v)f_v+\frac{1}{2}vf_{ss}+\rho\sigma
vf_{sv}+\frac{1}{2}\sigma^2vf_{vv}\right]-Ke^{-r(T-t)}\left[g_t+(r-\frac{1}{2}v)g_s\right.\\
& &\left.+(a-bv)g_v+\frac{1}{2}vg_{ss}+\rho\sigma
vg_{sv}+\frac{1}{2}\sigma^2vg_{vv}\right]+rsf-re^{-r(T-t)}Kg\\
&=&rc.
\end{eqnarray*}That is, $c$ satisfies the PDE (6.9.26).
\end{proof}

(iii)\begin{proof} First, by Markov property,
$f(t,X_t,V_t)=E[1_{\{X_T\ge \log K\}}|{\cal F}_t]$. So
$f(T,X_t,V_t)=1_{\{X_T\ge\log K\}}$, which implies
$f(T,x,v)=1_{\{x\ge \log K\}}$ for all $x\in\mathbb R$, $v\ge 0$.
Second, $f(t,X_t,V_t)$ is a martingale, so by differentiating $f$
and setting the $dt$ term as zero, we have the PDE (6.9.32) for $f$.
Indeed,
\begin{eqnarray*}
df(t,X_t,V_t)&=&\left[f_t(t,X_t,V_t)+f_x(t,X_t,V_t)(r+\frac{1}{2}V_t)+f_v(t,X_t,V_t)(a-bv_t+\rho\sigma
V_t)+\frac{1}{2}f_{xx}(t,X_t,V_t)V_t\right.\\
&
&\left.+\frac{1}{2}f_{vv}(t,X_t,V_t)\sigma^2V_t+f_{xv}(t,X_t,V_t)\sigma
V_t\rho\right]dt+\mbox{martingale part}.
\end{eqnarray*}
So we must have $f_t+(r+\frac{1}{2}v)f_x+(a-bv+\rho\sigma
v)f_v+\frac{1}{2}f_{xx}v+\frac{1}{2}f_{vv}\sigma^2v+\sigma v\rho
f_{xv}=0$. This is (6.9.32).

\end{proof}

(iv)\begin{proof} Similar to (iii).\end{proof}

(v) \begin{proof} $c(T,s,v)=sf(T,\log s,v)-e^{-r(T-t)}Kg(T,\log
s,v)=s1_{\{\log s\ge \log K\}}-K1_{\{\log s\ge \log K\}}=1_{\{s\ge
K\}}(s-K)=(s-K)^+$. \end{proof}

\noindent 6.8. \begin{proof} We follow the hint. Suppose $h$ is
smooth and compactly supported, then it is legitimate to exchange
integration and differentiation:
\begin{eqnarray*}
& &g_t(t,x)=\frac{\partial}{\partial
t}\int_0^{\infty}h(y)p(t,T,x,y)dy=\int_0^{\infty}h(y)p_t(t,T,x,y)dy,\\
& &g_x(t,x)=\int_0^{\infty}h(y)p_x(t,T,x,y)dy,\\
& &g_{xx}(t,x)=\int_0^{\infty}h(y)p_{xx}(t,T,x,y)dy.
\end{eqnarray*}
So (6.9.45) implies
$\int_0^{\infty}h(y)\left[p_t(t,T,x,y)+\beta(t,x)p_x(t,T,x,y)+\frac{1}{2}\gamma^2(t,x)p_{xx}(t,T,x,y)\right]dy=0$.
By the arbitrariness of $h$ and assuming $\beta$, $p_t$, $p_x$, $v$,
$p_{xx}$ are all continuous, we have
\[
p_t(t,T,x,y)+\beta(t,x)p_x(t,T,x,y)+\frac{1}{2}\gamma^2(t,x)p_{xx}(t,T,x,y)=0.
\]This is (6.9.43).
\end{proof}

\noindent 6.9.\begin{proof} We first note
\begin{eqnarray*}
dh_b(X_u)&=&h'_b(X_u)dX_u+\frac{1}{2}h_b''(X_u)dX_udX_u\\
&=&\left[h_b'(X_u)\beta(u,X_u)+\frac{1}{2}\gamma^2(u,X_u)h_b''(X_u)\right]du+h_b'(X_u)\gamma(u,X_u)dW_u.
\end{eqnarray*}
Integrate on both sides of the equation, we have
\[
h_b(X_T)-h_b(X_t)=\int_t^T\left[h_b'(X_u)\beta(u,X_u)+\frac{1}{2}\gamma^2(u,X_u)h_b''(X_u)\right]du+\mbox{martingale
part}.\] Take expectation on both sides, we get
\begin{eqnarray*}
E^{t,x}[h_b(X_T)-h_b(X_t)]
&=&\int_{-\infty}^{\infty}h_b(y)p(t,T,x,y)dy-h(x)\\
&=&\int_t^TE^{t,x}[h_b'(X_u)\beta(u,X_u)+\frac{1}{2}\gamma^2(u,X_u)h_b''(X_u)]du\\
&=&\int_t^T\int_{-\infty}^{\infty}\left[h_b'(y)\beta(u,y)+\frac{1}{2}\gamma^2(u,y)h_b''(y)\right]p(t,u,x,y)dydu.
\end{eqnarray*}
Since $h_b$ vanishes outside $(0,b)$, the integration range can be
changed from $(-\infty,\infty)$ to $(0,b)$, which gives (6.9.48).

By integration-by-parts formula, we have
\begin{eqnarray*}
\int_0^b\beta(u,y)p(t,u,x,y)h_b'(y)dy &=&
h_b(y)\beta(u,y)p(t,u,x,y)|_0^b-\int_0^bh_b(y)\frac{\partial}{\partial
y}(\beta(u,y)p(t,u,x,y))dy\\
&=&-\int_0^bh_b(y)\frac{\partial}{\partial
y}(\beta(u,y)p(t,u,x,y))dy,
\end{eqnarray*}
and
\begin{eqnarray*}
\int_0^b\gamma^2(u,y)p(t,u,x,y)h_b''(y)dy&=&-\int_0^b\frac{\partial}{\partial
y}(\gamma^2(u,y)p(t,u,x,y))h_b'(y)dy\\
&=&\int_0^b\frac{\partial^2}{\partial
y}(\gamma^2(u,y)p(t,u,x,y))h_b(y)dy.
\end{eqnarray*}
Plug these formulas into (6.9.48), we get (6.9.49).

Differentiate w.r.t. T on both sides of (6.9.49), we have
\[
\int_0^bh_b(y)\frac{\partial}{\partial
T}p(t,T,x,y)dy=-\int_0^b\frac{\partial}{\partial
y}[\beta(T,y)p(t,T,x,y)]h_b(y)dy+\frac{1}{2}\int_0^b\frac{\partial^2}{\partial
y^2}[\gamma^2(T,y)p(t,T,x,y)]h_b(y)dy,
\]that is, \[
\int_0^bh_b(y)\left[\frac{\partial}{\partial
T}p(t,T,x,y)+\frac{\partial }{\partial
y}(\beta(T,y)p(t,T,x,y))-\frac{1}{2}\frac{\partial^2}{\partial
y^2}(\gamma^2(T,y)p(t,T,x,y))\right]dy=0.
\]This is (6.9.50).

By (6.9.50) and the arbitrariness of $h_b$, we conclude for any
$y\in (0,\infty)$,
\[
\frac{\partial}{\partial T}p(t,T,x,y)+\frac{\partial}{\partial
y}(\beta(T,y)p(t,T,x,y))-\frac{1}{2}\frac{\partial^2}{\partial
y^2}(\gamma^2(T,y)p(t,T,x,y))=0.
\]
\end{proof}


\noindent 6.10. \begin{proof} Under the assumption that
$\lim_{y\to\infty}(y-K)ry\widetilde p(0,T,x,y)=0$, we have
\begin{eqnarray*}
-\int_K^{\infty}(y-K)\frac{\partial}{\partial y}(ry\widetilde
p(0,T,x,y))dy&=&-(y-K)ry\widetilde
p(0,T,x,y)|_K^{\infty}+\int_K^{\infty}ry\widetilde
p(0,T,x,y)dy\\
&=&\int_K^{\infty}ry\widetilde p(0,T,x,y)dy.
\end{eqnarray*}
If we further assume (6.9.57) and (6.9.58), then use
integration-by-parts formula twice, we have
\begin{eqnarray*}
& &\frac{1}{2}\int_K^{\infty}(y-K)\frac{\partial^2}{\partial
y^2}(\sigma^2(T,y)y^2\widetilde p(0,T,x,y))dy\\
&=&\frac{1}{2}\left[(y-K)\frac{\partial}{\partial
y}(\sigma^2(T,y)y^2\widetilde
p(0,T,x,y))|_K^{\infty}-\int_K^{\infty}\frac{\partial}{\partial
y}(\sigma^2(T,y)y^2\widetilde p(0,T,x,y))dy\right]\\
&=&-\frac{1}{2}(\sigma^2(T,y)y^2\widetilde p(0,T,x,y)|_K^{\infty})\\
&=&\frac{1}{2}\sigma^2(T,K)K^2\widetilde p(0,T,x,K).
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
c_T(0,T,x,K) &=&-rc(0,T,x,K)+e^{-rT}\int_K^{\infty}(y-K)\widetilde
p_T(0,T,x,y)dy\\
&=&-re^{-rT}\int_K^{\infty}(y-K)\widetilde
p(0,T,x,y)dy+e^{-rT}\int_K^{\infty}(y-K)\widetilde p_T(0,T,x,y)dy\\
&=&-re^{-rT}\int_K^{\infty}(y-K)\widetilde
p(0,T,x,y)dy-e^{-rT}\int_K^{\infty}(y-K)\frac{\partial}{\partial
y}(ry\widetilde
p(t,T,x,y))dy\\
& &+e^{-rT}\int_K^{\infty}(y-K)\frac{1}{2}\frac{\partial^2}{\partial
y^2}(\sigma^2(T,y)y^2\widetilde p(t,T,x,y))dy\\
&=&-re^{-rT}\int_K^{\infty}(y-K)\widetilde
p(0,T,x,y)dy+e^{-rT}\int_K^{\infty}ry\widetilde
p(0,T,x,y)dy\\
& &+e^{-rT}\frac{1}{2}\sigma^2(T,K)K^2\widetilde
p(0,T,x,K)\\
&=&re^{-rT}K\int_K^{\infty}\widetilde
p(0,T,x,y)dy+\frac{1}{2}e^{-rT}\sigma^2(T,K)K^2\widetilde
p(0,T,x,K)\\
&=&-rKc_K(0,T,x,K)+\frac{1}{2}\sigma^2(T,K)K^2c_{KK}(0,T,x,K).
\end{eqnarray*}

\end{proof}


\medskip

\section{Exotic Options}


$\bigstar$ {\bf Comments}:

\medskip

{\it On the PDE approach to pricing knock-out barrier options}. We give some clarification to the explanation below Theorem 7.3.1 (page 301-302), where the key is that $V(t)$ and $v(t,x)$ differ by an indicator function and all the hassles come from this difference.

More precisely, we define the first passage time $\rho$ by following the notation of the textbook:
\[
\rho = \inf\{t>0: S(t) = B\}.
\]
Then risk-neutral pricing gives the time-$t$ price of the knock-out barrier option as
\[
V(t) = \widetilde {\mathbb E} \left[\left.e^{-r(T-t)}V(T)\right|{\cal F}(t)\right]
\]
where $V(T) = (S(T)-K)^+ 1_{\{\rho \ge T\}} = (S(T)-K)^+ 1_{\{\rho > T\}}$, since the event $\{\rho=T\} \subset \{S(T)=B\}$ has zero probability.

Therefore, the discounted value process $e^{-rt}V(t) = \widetilde {\mathbb E} \left[\left.e^{-rT}V(T)\right|{\cal F}(t)\right]$ is a martingale under the risk-neutral measure $\widetilde {\mathbb P}$, but we cannot say $V(t)$ is solely a function of $t$ and $S(t)$, since it depends on the path property before $t$ as well. To see this analytically, we note
\[
\{\rho > T\} = \{\rho > t, \rho\circ\theta_t > T-t\},
\]
where $\theta_t$ is the {\it shift operator} used for sample paths (i.e. $\theta_t(\omega_{\cdot}) = \omega_{t+\cdot}$. See \O ksendal \cite[page~119]{Oksendal03} for details). Then
\[
V(t) = \widetilde {\mathbb E} \left[\left.e^{-r (T-t)}V(T)\right|{\cal F}(t)\right] = \widetilde {\mathbb E} \left[\left.e^{-r (T-t)}(S(T)-K)^+1_{\{\rho\circ\theta_t > T-t\}}\right|{\cal F}(t)\right] 1_{\{\rho > t\}}.
\]
Note $\rho\circ\theta_t$ is solely dependent on the behavior of sample paths between time $t$ and $T$. Therefore, we can apply Markov property
\[
V(t) = e^{-r(T-t)} \widetilde {\mathbb E}^{S(t)}[(S(T-t)-K)^+1_{\{\rho > T-t\}}] 1_{\{\rho > t\}}
\]

Define $v(t,x) = e^{-r(T-t)} \widetilde {\mathbb E}^{x}[(S(T-t)-K)^+1_{\{\rho > T-t\}}]$, then $V(t) = v(t,x) 1_{\{\rho > t\}}$ and $v(t,x)$ satisfies the conditions (7.3.4)-(7.3.7) listed in Theorem 7.3.1. Indeed, it is easy to see $v(t,x)$ satisfies all the boundary conditions (7.3.5)-(7.3.7), by the arguments in the paragraph immediately after Theorem 7.3.1. For the Black-Scholes-Merton PDE (7.3.4), we note
\begin{eqnarray*}
d[e^{-rt}V(t)] &=& d[e^{-rt}v(t,S(t))1_{\{\rho>t\}}] = 1_{\{\rho>t\}} d[e^{-rt}v(t,S(t))] + e^{-rt}v(t,S(t)) d 1_{\{\rho>t\}}.
\end{eqnarray*}
Following the computation in equation (7.3.13), We can see the Black-Scholes-Merton equation (7.3.4) must hold for $\{(t,x): 0 \le t < T, 0 \le x < B\}$.
\bigskip

\noindent 7.1. (i) \begin{proof} Since
$\delta_{\pm}(\tau,s)=\frac{1}{\sigma\sqrt{\tau}}[\log
s+(r\pm\frac{1}{2}\sigma^2)\tau]=\frac{\log
s}{\sigma}\tau^{-\frac{1}{2}}+\frac{r\pm\frac{1}{2}\sigma^2}{\sigma}\sqrt{\tau}$,
\begin{eqnarray*}
\frac{\partial}{\partial t}\delta_{\pm}(\tau,s) &=&\frac{\log
s}{\sigma}(-\frac{1}{2})\tau^{-\frac{3}{2}}\frac{\partial
\tau}{\partial
t}+\frac{r\pm\frac{1}{2}\sigma^2}{\sigma}\frac{1}{2}\tau^{-\frac{1}{2}}\frac{\partial\tau}{\partial
t}\\
&=&-\frac{1}{2\tau}\left[\frac{\log
s}{\sigma}\frac{1}{\sqrt{\tau}}(-1)-\frac{r\pm\frac{1}{2}\sigma^2}{\sigma}\sqrt{\tau}(-1)\right]\\
&=&-\frac{1}{2\tau}\cdot\frac{1}{\sigma\sqrt{\tau}}\left[-\log s
s+(r\pm\frac{1}{2}\sigma^2)\tau)\right]\\
&=&-\frac{1}{2\tau}\delta_{\pm}(\tau,\frac{1}{s}).
\end{eqnarray*}
\end{proof}

(ii)\begin{proof}
\[
\frac{\partial}{\partial
x}\delta_{\pm}(\tau,\frac{x}{c})=\frac{\partial }{\partial
x}\left(\frac{1}{\sigma\sqrt{\tau}}\left[\log
\frac{x}{c}+(r\pm\frac{1}{2}\sigma^2)\tau\right]\right)=\frac{1}{x\sigma\sqrt{\tau}},
\]
\[
\frac{\partial}{\partial
x}\delta_{\pm}(\tau,\frac{c}{x})=\frac{\partial }{\partial
x}\left(\frac{1}{\sigma\sqrt{\tau}}\left[\log
\frac{c}{x}+(r\pm\frac{1}{2}\sigma^2)\tau\right]\right)=-\frac{1}{x\sigma\sqrt{\tau}}.
\]
\end{proof}

(iii)\begin{proof}
\[
N'(\delta_{\pm}(\tau,s))=\frac{1}{\sqrt{2\pi}}e^{-\frac{\delta_{\pm}(\tau,s)}{2}}=\frac{1}{\sqrt{2\pi}}e^{-\frac{(\log
s+r\tau)^2\pm\sigma^2\tau(\log
s+r\tau)+\frac{1}{4}\sigma^4\tau^2}{2\sigma^2\tau}}.
\]Therefore
\[
\frac{N'(\delta_+(\tau,s))}{N'(\delta_-(\tau,s))}=e^{-\frac{2\sigma^2\tau(\log
s+r\tau)}{2\sigma^2\tau}}=\frac{e^{-r\tau}}{s}
\]and
$e^{-r\tau}N'(\delta_-(\tau,s))=sN'(\delta_+(\tau,s))$.
\end{proof}

(iv)\begin{proof}
\[
\frac{N'(\delta_{\pm}(\tau,s))}{N'(\delta_{\pm}(\tau,s^{-1}))}=e^{-\frac{\left[(\log
s+r\tau)^2-(\log\frac{1}{s}+r\tau)^2\right]\pm\sigma^2\tau(\log
s-\log\frac{1}{s})}{2\sigma^2\tau}}=e^{-\frac{4r\tau\log s\pm
2\sigma^2\tau\log s}{2\sigma^2\tau}}=e^{-(\frac{2r}{\sigma^2}\pm
1)\log s}=s^{-(\frac{2r}{\sigma^2}\pm 1)}.
\]So $N'(\delta_{\pm}(\tau,s^{-1}))=s^{(\frac{2r}{\sigma^2}\pm
1)}N'(\delta_{\pm}(\tau,s))$.
\end{proof}

(v)\begin{proof} $
\delta_+(\tau,s)-\delta_-(\tau,s)=\frac{1}{\sigma\sqrt{\tau}}\left[\log
s+(r+\frac{1}{2}\sigma^2)\tau\right]-\frac{1}{\sigma\sqrt{\tau}}\left[\log
s+(r-\frac{1}{2}\sigma^2)\tau\right]=\frac{1}{\sigma\sqrt{\tau}}\sigma^2\tau=\sigma\sqrt{\tau}.
$
\end{proof}

(vi)\begin{proof} $
\delta_{\pm}(\tau,s)-\delta_{\pm}(\tau,s^{-1})=\frac{1}{\sigma\sqrt{\tau}}\left[\log
s+(r\pm\frac{1}{2}\sigma^2)\tau\right]-\frac{1}{\sigma\sqrt{\tau}}\left[\log
s^{-1}+(r \pm \frac{1}{2}\sigma^2)\tau\right]=\frac{2\log
s}{\sigma\sqrt{\tau}}. $
\end{proof}

(vii) \begin{proof} $N'(y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}$,
so
$N''(y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}(-\frac{y^2}{2})'=-yN'(y)$.
\end{proof}

{\bf To be continued ...}


\noindent 7.3. \begin{proof} We note $S_T=S_0e^{\sigma\widehat
W_T}=S_te^{\sigma(\widehat W_T-\widehat W_t)}$, $\widehat
W_T-\widehat W_t=(\widetilde W_T-\widetilde W_t)+\alpha(T-t)$ is
independent of ${\cal F}_t$, $\sup_{t\le u\le T}(\widehat
W_u-\widehat W_t)$ is independent of ${\cal F}_t$, and
\begin{eqnarray*}
Y_T
&=&S_0e^{\sigma\widehat M_T}\\
&=&S_0e^{\sigma\sup_{t\le u\le T}\widehat W_u}1_{\{\widehat M_t\le
\sup_{t\le u\le T}\widehat W_t\}}+S_0e^{\sigma \widehat
M_t}1_{\{\widehat M_t>\sup_{t\le u\le T}\widehat W_u\}}\\
&=&S_te^{\sigma\sup_{t\le u\le T}(\widehat W_u-\widehat
W_t)}1_{\{\frac{Y_t}{S_t}\le e^{\sigma\sup_{t\le u\le T}(\widehat
W_u-\widehat W_t)} \}}+Y_t1_{\{\frac{Y_t}{S_t}\le
e^{\sigma\sup_{t\le u\le T}(\widehat W_u-\widehat W_t)}\}}.
\end{eqnarray*}
So $E[f(S_T,Y_T)|{\cal
F}_t]=E[f(x\frac{S_{T-t}}{S_0},x\frac{Y_{T-t}}{S_0}1_{\{\frac{y}{x}\le\frac{Y_{T-t}}{S_0}\}}+y1_{\{\frac{y}{x}\le
\frac{Y_{T-t}}{S_0}\}})]$, where $x=S_t$, $y=Y_t$. Therefore
$E[f(S_T,Y_T)|{\cal F}_t]$ is a Borel function of $(S_t,Y_t)$.

\end{proof}


\noindent 7.4. \begin{proof} By Cauchy's inequality and the
monotonicity of $Y$, we have
\begin{eqnarray*}
|\sum_{j=1}^m(Y_{t_j}-Y_{t_{j-1}})(S_{t_j}-S_{t_{j-1}})| &\le &
\sum_{j=1}^m|Y_{t_j}-Y_{t_{j-1}}||S_{t_j}-S_{t_{j-1}}|\\
&\le&
\sqrt{\sum_{j=1}^m(Y_{t_j}-Y_{t_{j-1}})^2}\sqrt{\sum_{j=1}^m(S_{t_j}-S_{t_{j-1}})^2}\\
&\le&\sqrt{\max_{1\le j\le
m}|Y_{t_j}-Y_{t_{j-1}}|(Y_T-Y_0)}\sqrt{\sum_{j=1}^m(S_{t_j}-S_{t_{j-1}})^2}.
\end{eqnarray*}
If we increase the number of partition points to infinity and let
the length of the longest subinterval $\max_{1\le j\le
m}|t_j-t_{j-1}|$ approach zero, then
$\sqrt{\sum_{j=1}^m(S_{t_j}-S_{t_{j-1}})^2}\to\sqrt{[S]_T-[S]_0}<\infty$
and $\max_{1\le j\le m}|Y_{t_j}-Y_{t_{j-1}}|\to 0$ a.s. by the
continuity of $Y$. This implies
$\sum_{j=1}^m(Y_{t_j}-Y_{t_{j-1}})(S_{t_j}-S_{t_{j-1}})\to 0$.
\end{proof}


\medskip

\section{American Derivative Securities}


$\bigstar$ {\bf Comments}:

\medskip
{\it Justification of Definition 8.3.1}. For any given stopping time $\tau$, let $C_t = \int_0^t (K-S(u))d 1_{\{\tau \le u\}}$, then
\[
e^{-r\tau} (K-S(\tau))1_{\{\tau<\infty\}} = \int_0^{\infty} e^{-rt} dC_t.
\]
Regarding $C$ as a cumulative cash flow, valuation of put option via (8.3.2) is justified by the risk-neutral valuation of a cash flow via (5.6.10).

\bigskip

\noindent 8.1. \begin{proof}
$v_L'(L+)=\left.(K-L)(-\frac{2r}{\sigma^2})(\frac{x}{L})^{-\frac{2r}{\sigma^2}-1}\frac{1}{L}\right|_{x=L}=-\frac{2r}{\sigma^2L}(K-L)$.
So $v_L'(L+)=v_L'(L-)$ if and only if
$-\frac{2r}{\sigma^2L}(K-L)=-1$. Solve for $L$, we get
$L=\frac{2rK}{2r+\sigma^2}$.
\end{proof}

\noindent 8.2. \begin{proof} By the calculation in Section 8.3.3, we
can see $v_2(x)\ge (K_2-x)^+\ge (K_1-x)^+$,
$rv_2(x)-rxv_2'(x)-\frac{1}{2}\sigma^2x^2v_2''(x)\ge 0$ for all
$x\ge 0$, and for $0\le x<L_{1*}<L_{2*}$,
\[
rv_2(x)-rxv_2'(x)-\frac{1}{2}\sigma^2x^2v_2''(x)=rK_2>rK_1>0.
\]So the linear complementarity conditions for $v_2$ imply
$v_2(x)=(K_2-x)^+=K_2-x>K_1-x=(K_1-x)^+$ on $[0,L_{1*}]$. Hence
$v_2(x)$ does not satisfy the third linear complementarity condition
for $v_1$: for each $x\ge 0$, equality holds in either (8.8.1) or
(8.8.2) or both.
\end{proof}

\noindent 8.3. (i)\begin{proof} Suppose $x$ takes its values in a
domain bounded away from $0$. By the general theory of linear
differential equations, if we can find two linearly independent
solutions $v_1(x)$, $v_2(x)$ of (8.8.4), then any solution of
(8.8.4) can be represented in the form of $C_1v_1+C_2v_2$ where
$C_1$ and $C_2$ are  constants. So it suffices to find two linearly
independent special solutions of (8.8.4). Assume $v(x)=x^p$ for some
constant $p$ to be determined, (8.8.4) yields
$x^p(r-pr-\frac{1}{2}\sigma^2p(p-1))=0$. Solve the quadratic
equation
$0=r-pr-\frac{1}{2}\sigma^2p(p-1)=(-\frac{1}{2}\sigma^2p-r)(p-1)$,
we get $p=1$ or $-\frac{2r}{\sigma^2}$. So a general solution of
(8.8.4) has the form $C_1x+C_2x^{-\frac{2r}{\sigma^2}}$.
\end{proof}

(ii) \begin{proof} Assume there is an interval $[x_1,x_2]$ where
$0<x_1<x_2<\infty$, such that $v(x)\not \equiv 0$ satisfies (8.3.19)
with equality on $[x_1,x_2]$ and satisfies (8.3.18) with equality
for x at and immediately to the left of $x_1$ and for $x$ at and
immediately to the right of $x_2$, then we can find some $C_1$ and
$C_2$, so that $v(x)=C_1x+C_2x^{-\frac{2r}{\sigma^2}}$ on
$[x_1,x_2]$. If for some $x_0\in [x_1,x_2]$, $v(x_0)=v'(x_0)=0$, by
the uniqueness of the solution of (8.8.4), we would conclude
$v\equiv 0$. This is a contradiction. So such an $x_0$ cannot exist.
This implies $0<x_1<x_2<K$ (if $K\le x_2$, $v(x_2)=(K-x_2)^+=0$ and
$v'(x_2)$=the right derivative of $(K-x)^+$ at $x_2$, which is 0).
\footnote{Note we have interpreted the condition ``$v(x)$ satisfies
(8.3.18) with equality for $x$ at and immediately to the right of
$x_2$'' as ``$v(x_2)=(K-x_2)^+$ and $v'(x_2)=$the right derivative
of $(K-x)^+$ at $x_2$.'' This is weaker than ``$v(x)=(K-x)$ in a
right neighborhood of $x_2$."} Thus we have four equations for $C_1$
and $C_2$:
\begin{eqnarray*}
\begin{cases}
C_1x_1+C_2x_1^{-\frac{2r}{\sigma^2}}=K-x_1\\
C_1x_2+C_2x_2^{-\frac{2r}{\sigma^2}}=K-x_2\\
C_1-\frac{2r}{\sigma^2}C_2x_1^{-\frac{2r}{\sigma^2}-1}=-1\\
C_1-\frac{2r}{\sigma^2}C_2x_2^{-\frac{2r}{\sigma^2}-1}=-1.
\end{cases}
\end{eqnarray*}
Since $x_1\ne x_2$, the last two equations imply $C_2=0$. Plug
$C_2=0$ into the first two equations, we have
$C_1=\frac{K-x_1}{x_1}=\frac{K-x_2}{x_2}$; plug $C_2=0$ into the
last two equations, we have $C_1=-1$. Combined, we would have
$x_1=x_2$. Contradiction. Therefore our initial assumption is
incorrect, and the only solution $v$ that satisfies the specified
conditions in the problem is the zero solution.
\end{proof}

(iii)\begin{proof} If in a right neighborhood of $0$, $v$ satisfies
(8.3.19) with equality, then part (i) implies
$v(x)=C_1x+C_2x^{-\frac{2r}{\sigma^2}}$ for some constants $C_1$ and
$C_2$. Then $v(0)=\lim_{x\downarrow 0}v(x)=0<(K-0)^+$, i.e. (8.3.18)
will be violated. So we must have
$rv-rxv'-\frac{1}{2}\sigma^2x^2v''>0$ in a right neighborhood of
$0$. According to (8.3.20), $v(x)=(K-x)^+$ near o. So $v(0)=K$. We
have thus concluded simultaneously that $v$ cannot satisfy (8.3.19)
with equality near 0 and $v(0)=K$, starting from first principles
(8.3.18)-(8.3.20).
\end{proof}

(iv)\begin{proof}This is already shown in our solution of part
(iii): near 0, $v$ cannot satisfy (8.3.19) with equality.\end{proof}

(v)\begin{proof} If $v$ satisfy $(K-x)^+$ with equality for all
$x\ge 0$, then $v$ cannot have a continuous derivative as stated in
the problem. This is a contradiction.
\end{proof}

(vi)\begin{proof} By the result of part (i), we can start with
$v(x)=(K-x)^+$ on $[0,x_1]$ and
$v(x)=C_1x+C_2x^{-\frac{2r}{\sigma^2}}$ on $[x_1,\infty)$. By the
assumption of the problem, both $v$ and $v'$ are continuous. Since
$(K-x)^+$ is not differentiable at $K$, we must have $x_1\le K$.This
gives us the equations
\begin{eqnarray*}
\begin{cases}
K-x_1=(K-x_1)^+=C_1x_1+C_2x_1^{-\frac{2r}{\sigma^2}}\\
-1=C_1-\frac{2r}{\sigma^2}C_2x_1^{-\frac{2r}{\sigma^2}-1}.
\end{cases}
\end{eqnarray*}
Because $v$ is assumed to be bounded, we must have $C_1=0$ and the
above equations only have two unknowns: $C_2$ and $x_1$. Solve them
for $C_2$ and $x_1$, we are done.
\end{proof}

\noindent 8.4. (i) \begin{proof} This is already shown in part (i)
of Exercise 8.3.\end{proof}

(ii)\begin{proof} We solve for $A$, $B$ the equations
\begin{eqnarray*}
\begin{cases}
AL^{-\frac{2r}{\sigma^2}}+BL=K-L\\
-\frac{2r}{\sigma^2}AL^{-\frac{2r}{\sigma^2}-1}+B=-1,
\end{cases}
\end{eqnarray*}
and we obtain
$A=\frac{\sigma^2KL^{\frac{2r}{\sigma^2}}}{\sigma^2+2r}$,
$B=\frac{2rK}{L(\sigma^2+2r)}-1$.
\end{proof}

(iii) \begin{proof} By (8.8.5), $B>0$. So for $x\ge K$, $f(x)\ge
BK>0=(K-x)^+$. If $L\le x<K$,
\begin{eqnarray*}
f(x)-(K-x)^+
&=&
\frac{\sigma^2KL^{\frac{2r}{\sigma^2}}}{\sigma^2+2r}x^{-\frac{2r}{\sigma^2}}+\frac{2rKx}{L(\sigma^2+2r)}-K \\
&=& x^{-\frac{2r}{\sigma^2}}\frac{KL^{\frac{2r}{\sigma^2}}\left[\sigma^2+2r(\frac{x}{L})^{\frac{2r}{\sigma^2}+1}-(\sigma^2+2r)(\frac{x}{L})^{\frac{2r}{\sigma^2}}\right]}{(\sigma^2+2r)L}.
\end{eqnarray*}
Let
$g(\theta)=\sigma^2+2r\theta^{\frac{2r}{\sigma^2}+1}-(\sigma^2+2r)\theta^{\frac{2r}{\sigma^2}}$
with $\theta\ge 1$. Then $g(1)=0$ and
$g'(\theta)=2r(\frac{2r}{\sigma^2}+1)\theta^{\frac{2r}{\sigma^2}}-(\sigma^2+2r)\frac{2r}{\sigma^2}\theta^{\frac{2r}{\sigma^2}-1}=\frac{2r}{\sigma^2}(\sigma^2+2r)\theta^{\frac{2r}{\sigma^2}-1}(\theta-1)\ge
0$. So $g(\theta)\ge 0$ for any $\theta\ge 1$. This shows $f(x)\ge
(K-x)^+$ for $L\le x<K$. Combined, we get $f(x)\ge (K-x)^+$ for all
$x\ge L$.
\end{proof}

(iv)\begin{proof} Since
$\lim_{x\to\infty}v(x)=\lim_{x\to\infty}f(x)=\infty$ and
$\lim_{x\to\infty}v_{L_*}(x)=\lim_{x\to\infty}(K-L_*)(\frac{x}{L_*})^{-\frac{2r}{\sigma^2}}=0$,
$v(x)$ and $v_{L_*}(x)$ are different. By part (iii), $v(x)\ge
(K-x)^+$. So $v$ satisfies (8.3.18). For $x\ge L$,
$rv-rxv'-\frac{1}{2}\sigma^2x^2v''=rf-rxf-\frac{1}{2}\sigma^2x^2f''=0$.
For $0\le x\le L$, $rv-rxv'-\frac{1}{2}\sigma^2x^2v''=r(K-x)+rx=rK$.
Combined, $rv-rxv'-\frac{1}{2}\sigma^2x^2v''\ge 0$ for $x\ge 0$. So
$v$ satisfies (8.3.19). Along the way, we also showed $v$ satisfies
(8.3.20). In summary, $v$ satisfies the linear complementarity
condition (8.3.18)-(8.3.20), but $v$ is not the function $v_{L_*}$
given by (8.3.13).
\end{proof}

(v)\begin{proof}By part (ii), $B=0$ if and only if
$\frac{2rK}{L(\sigma^2+2r)}-1=0$, i.e. $L=\frac{2rK}{2r+\sigma^2}$.
In this case,
$v(x)=Ax^{-\frac{2r}{\sigma^2}}=\frac{\sigma^2K}{\sigma^2+2r}(\frac{x}{L})^{-\frac{2r}{\sigma^2}}=(K-L)(\frac{x}{L})^{-\frac{2r}{\sigma^2}}=v_{L_*}(x)$,
on the interval $[L,\infty)$.\end{proof}

\noindent 8.5. The difficulty of the dividend-paying case is that
from Lemma 8.3.4, we can only obtain $\widetilde
E[e^{-(r-a)\tau_L}]$, not $\widetilde E[e^{-r\tau_L}]$. So we have
to start from Theorem 8.3.2.

(i)\begin{proof} By (8.8.9), $S_t=S_0e^{\sigma\widetilde
W_t+(r-a-\frac{1}{2}\sigma^2)t}$. Assume $S_0=x$, then $S_t=L$ if
and only if $-\widetilde
W_t-\frac{1}{\sigma}(r-a-\frac{1}{2}\sigma^2)t=\frac{1}{\sigma}\log\frac{x}{L}$.
By Theorem 8.3.2,
\[
\widetilde
E[e^{-r\tau_L}]=e^{-\frac{1}{\sigma}\log\frac{x}{L}\left[\frac{1}{\sigma}(r-a-\frac{1}{2}\sigma^2)+\sqrt{\frac{1}{\sigma^2}(r-a-\frac{1}{2}\sigma^2)^2+2r}\right]}.
\]If we set
$\gamma=\frac{1}{\sigma^2}(r-a-\frac{1}{2}\sigma^2)+\frac{1}{\sigma}\sqrt{\frac{1}{\sigma^2}(r-a-\frac{1}{\sigma^2})^2+2r}$,
we can write $\widetilde E[e^{-r\tau_L}]$ as
$e^{-\gamma\log\frac{x}{L}}=(\frac{x}{L})^{-\gamma}$. So the
risk-neutral expected discounted pay off of this strategy is
\begin{eqnarray*}
v_L(x)=
\begin{cases}
K-x, & 0\le x\le L\\
(K-L)(\frac{x}{L})^{-\gamma}, & x>L.
\end{cases}
\end{eqnarray*}
\end{proof}

(ii)\begin{proof} $\frac{\partial}{\partial
L}v_L(x)=-(\frac{x}{L})^{-\gamma}(1-\frac{\gamma(K-L)}{L})$. Set
$\frac{\partial}{\partial L}v_L(x)=0$ and solve for $L_*$, we have
$L_*=\frac{\gamma K}{\gamma+1}$.
\end{proof}

(iii)\begin{proof} By It\^{o}'s formula, we have
\[
d\left[e^{-rt}v_{L_*}(S_t)\right]=e^{-rt}\left[-rv_{L_*}(S_t)+v'_{L_*}(S_t)(r-a)S_t+\frac{1}{2}v_{L_*}''(S_t)\sigma^2S_t^2\right]dt+e^{-rt}v_{L_*}'(S_t)\sigma
S_t d\widetilde W_t.
\]
If $x>L_*$,
\begin{eqnarray*}
&
&-rv_{L_*}(x)+v_{L_*}'(x)(r-a)x+\frac{1}{2}v_{L_*}''(x)\sigma^2x^2\\
&=&-r(K-L_*)\left(\frac{x}{L_*}\right)^{-\gamma}+(r-a)x(K-L_*)(-\gamma)\frac{x^{-\gamma-1}}{L_*^{-\gamma}}+\frac{1}{2}\sigma^2x^2(-\gamma)(-\gamma-1)(K-L_*)\frac{x^{-\gamma-2}}{L^{-\gamma}_*}\\
&=&(K-L_*)\left(\frac{x}{L_*}\right)^{-\gamma}\left[-r-(r-a)\gamma+\frac{1}{2}\sigma^2\gamma(\gamma+1)\right].
\end{eqnarray*}
By the definition of $\gamma$, if we define
$u=r-a-\frac{1}{2}\sigma^2$, we have
\begin{eqnarray*}
& &r+(r-a)\gamma-\frac{1}{2}\sigma^2\gamma(\gamma+1)\\
&=&r-\frac{1}{2}\sigma^2\gamma^2+\gamma(r-a-\frac{1}{2}\sigma^2)\\
&=&r-\frac{1}{2}\sigma^2\left(\frac{u}{\sigma^2}+\frac{1}{\sigma}\sqrt{\frac{u^2}{\sigma^2}+2r}\right)^2+\left(\frac{u}{\sigma^2}+\frac{1}{\sigma}\sqrt{\frac{u^2}{\sigma^2}+2r}\right)u\\
&=&r-\frac{1}{2}\sigma^2\left(\frac{u^2}{\sigma^4}+\frac{2u}{\sigma^3}\sqrt{\frac{u^2}{\sigma^2}+2r}+\frac{1}{\sigma^2}\left(\frac{u^2}{\sigma^2}+2r\right)\right)+\frac{u^2}{\sigma^2}+\frac{u}{\sigma}\sqrt{\frac{u^2}{\sigma^2}+2r}\\
&=&r-\frac{u^2}{2\sigma^2}-\frac{u}{\sigma}\sqrt{\frac{u^2}{\sigma^2}+2r}-\frac{1}{2}\left(\frac{u^2}{\sigma^2}+2r\right)+\frac{u^2}{\sigma^2}+\frac{u}{\sigma}\sqrt{\frac{u^2}{\sigma^2}+2r}\\
&=&0.
\end{eqnarray*}
If $x<L_*$,
$-rv_{L_*}(x)+v'_{L_*}(x)(r-a)x+\frac{1}{2}v''_{L_*}(x)\sigma^2x^2=-r(K-x)+(-1)(r-a)x=-rK+ax$.
Combined, we get
\[
d\left[e^{-rt}v_{L_*}(S_t)\right]=-e^{-rt}1_{\{S_t<L_*\}}(rK-aS_t)dt+e^{-rt}v'_{L_*}(S_t)\sigma
S_td\widetilde W_t.
\]Following the reasoning in the proof of Theorem 8.3.5, we only
need to show $1_{\{x<L_*\}}(rK-ax)\ge 0$ to finish the solution.
This is further equivalent to proving $rK-aL_*\ge 0$. Plug
$L_*=\frac{\gamma K}{\gamma+1}$ into the expression and note
$\gamma\ge
\frac{1}{\sigma}\sqrt{\frac{1}{\sigma^2}(r-a-\frac{1}{2}\sigma^2)^2}+\frac{1}{\sigma^2}(r-a-\frac{1}{2}\sigma^2)\ge
0$, the inequality is further reduced to $r(\gamma+1)-a\gamma\ge 0$.
We prove this inequality as follows.

Assume for some $K$, $r$, $a$ and $\sigma$ ($K$ and $\sigma$ are
assumed to be strictly positive, $r$ and $a$ are assumed to be
non-negative), $rK-aL_*<0$, then necessarily $r<a$, since
$L_*=\frac{\gamma K}{\gamma+1}\le K$. As shown before, this means
$r(\gamma+1)-a\gamma< 0$. Define $\theta=\frac{r-a}{\sigma}$, then
$\theta<0$ and
$\gamma=\frac{1}{\sigma^2}(r-a-\frac{1}{2}\sigma^2)+\frac{1}{\sigma}\sqrt{\frac{1}{\sigma^2}(r-a-\frac{1}{2}\sigma^2)^2+2r}=\frac{1}{\sigma}(\theta-\frac{1}{2}\sigma)+\frac{1}{\sigma}\sqrt{(\theta-\frac{1}{2}\sigma)^2+2r}$.
We have
\begin{eqnarray*}
r(\gamma+1)-a\gamma< 0
&\iff& (r-a)\gamma+r<0 \\
&\iff& (r-a)\left[\frac{1}{\sigma}(\theta-\frac{1}{2}\sigma)+\frac{1}{\sigma}\sqrt{(\theta-\frac{1}{2}\sigma)^2+2r}\right]+r<0\\
&\iff&\theta(\theta-\frac{1}{2}\sigma)+\theta\sqrt{(\theta-\frac{1}{2}\sigma)^2+2r}+r<0\\
&\iff&\theta\sqrt{(\theta-\frac{1}{2}\sigma)^2+2r}<-r-\theta(\theta-\frac{1}{2}\sigma)(<0)\\
&\iff&\theta^2[(\theta-\frac{1}{2}\sigma)^2+2r]>
r^2+\theta^2(\theta-\frac{1}{2}\sigma)^2+2\theta
r(\theta-\frac{1}{2}\sigma^2)\\
&\iff&0>r^2-\theta r\sigma^2\\
&\iff&0>r-\theta\sigma^2.
\end{eqnarray*}
Since $\theta\sigma^2<0$, we have obtained a contradiction. So our
initial assumption is incorrect, and $rK-aL_*\ge 0$ must be true.
\end{proof}

(iv)\begin{proof} The proof is similar to that of Corollary 8.3.6.
Note the only properties used in the proof of Corollary 8.3.6 are
that $e^{-rt}v_{L_*}(S_t)$ is a supermartingale,
$e^{-rt\wedge\tau_{L_*}}v_{L_*}(S_t\wedge\tau_{L_*})$ is a
martingale, and $v_{L_*}(x)\ge (K-x)^+$. Part (iii) already proved
the supermartingale-martingale property, so it suffices to show
$v_{L_*}(x)\ge (K-x)^+$ in our problem. Indeed, by $\gamma\ge 0$,
$L_*=\frac{\gamma K}{\gamma+1}<K$. For $x\ge K>L_*$,
$v_{L_*}(x)>0=(K-x)^+$; for $0\le x<L_*$, $v_{L_*}(x)=K-x=(K-x)^+$;
finally, for $L_*\le x\le K$,
\[
\frac{d}{dx}(v_{L_*}(x)-(K-x))=-\gamma(K-L_*)\frac{x^{-\gamma-1}}{L_*^{-\gamma}}+1\ge
-\gamma(K-L_*)\frac{L_*^{-\gamma-1}}{L_*^{-\gamma}}+1=-\gamma(K-\frac{\gamma
K}{\gamma+1})\frac{1}{\frac{\gamma K}{\gamma+1}}+1=0.
\]and $(v_{L_*}(x)-(K-x))|_{x=L_*}=0$. So for $L_*\le x\le K$,
$v_{L_*}(x)-(K-x)^+\ge 0$. Combined, we have $v_{L_*}(x)\ge
(K-x)^+\ge 0$ for all $x\ge 0$.
\end{proof}

\noindent 8.6. \begin{proof} By Lemma 8.5.1, $X_t=e^{-rt}(S_t-K)^+$
is a submartingale. For any $\tau\in {\Gamma}_{0,T}$, Theorem 8.8.1
implies
\begin{eqnarray*}
\widetilde E[e^{-rT}(S_T-K)^+]\ge \widetilde E[e^{-r\tau\wedge
T}(S_{\tau\wedge T}-K)^+]\ge
E[e^{-r\tau}(S_{\tau}-K)^+1_{\{\tau<\infty\}}]=E[e^{-r\tau}(S_{\tau}-K)^+],
\end{eqnarray*}
where we take the convention that $e^{-r\tau}(S_{\tau}-K)^+= 0$ when
$\tau=\infty$. Since $\tau$ is arbitrarily chosen, $\widetilde
E[e^{-rT}(S_T-K)^+]\ge \max_{\tau\in \Gamma_{0,T}}\widetilde
E[e^{-r\tau}(S_{\tau}-K)^+]$. The other direction ``$\le$" is
trivial since $T\in \Gamma_{0,T}$.
\end{proof}

\noindent 8.7. \begin{proof} Suppose $\lambda\in [0,1]$ and $0\le
x_1\le x_2$, we have $f((1-\lambda)x_1+\lambda x_2)\le
(1-\lambda)f(x_1)+\lambda f(x_2)\le (1-\lambda)h(x_1)+\lambda
h(x_2)$. Similarly,  $g((1-\lambda)x_1+\lambda x_2)\le
(1-\lambda)h(x_1)+\lambda h(x_2)$. So
\[
h((1-\lambda)x_1+\lambda x_2)=\max\{f((1-\lambda)x_1+\lambda x_2),
g((1-\lambda)x_1+\lambda x_2)\}\le (1-\lambda)h(x_1)+\lambda h(x_2).
\]That is, $h$ is also convex.
\end{proof}



\medskip

\section{Change of Num\'{e}raire}


$\bigstar$ {\bf Comments}:

\medskip

1) To provide an intuition for change of num\'{e}raire, we give {\it a
summary of results for change of num\'{e}raire in discrete case}.
This summary is based on Shiryaev \cite{Shiryaev99}.

Consider a model of financial market $(\widetilde B,\bar B, {\bf S})$ as
in Delbaen and Schachermayer \cite{DS06} Definition 2.1.1 or Shiryaev \cite[page~383]{Shiryaev99}. Here
$\widetilde B$ and $\bar B$ are both one-dimensional while ${\bf S}$ could
be a vector price process. Suppose $\widetilde B$ and $\bar B$ are
both strictly positive, then both of them can be chosen as
num\'{e}aire.

Several results hold under this model. First, no-arbitrage and
completeness properties of market are independent of the choice of
num\'{e}raire (see, for example, Shiryaev \cite[page~413, 481]{Shiryaev99}).

Second, if the market is arbitrage-free, then
corresponding to $\widetilde B$ (resp. $\bar B$), there is an
equivalent probability measure $\widetilde {\mathbb P}$ (resp. $\bar {\mathbb P}$), such that
$\left(\frac{\bar B}{\widetilde B}, \frac{\bf S}{\widetilde B}\right)$
 (resp. $\left(\frac{\widetilde  B}{\bar B}, \frac{\bf S}{\bar
 B}\right)$) is a martingale under $\widetilde {\mathbb P}$ (resp. $\bar {\mathbb P}$).

Third, if the market is both arbitrage-free and complete, we have
the relation
\begin{eqnarray}\label{Radon-Nikodym density for numeraire}
d\bar {\mathbb P}=\frac{\bar B_T}{\widetilde B_T}\frac{1}{{\mathbb E}\left[\frac{\bar
B_0}{\widetilde B_0}\right]}d\widetilde {\mathbb P}.
\end{eqnarray} See Shiryaev \cite[page~510]{Shiryaev99}, formula (12).

Finally, if $f_T$ is a European contingent claim with maturity $N$ and the market
is both arbitrage-free and complete, then \footnote{If the market is
incomplete but the contingent claim is still replicable, this result
still holds for $t=0$. Indeed, in Shiryaev \cite{Shiryaev99},
Chapter V \S1c.2, it is easy to generalize formula (12) to the case
of $N\ge 1$ with $x^*=\sup_{\widetilde {\mathbb P} \in {\cal P}({\mathbb P})} \widetilde
{\mathbb E}\left[\frac{f_N}{B_N}\right]B_0$, $x_*=\inf_{\widetilde {\mathbb P} \in {\cal
P}({\mathbb P})}\widetilde {\mathbb E}\left[\frac{f_N}{B_N}\right]B_0$,
$C^*({\mathbb P})=\inf\{x\ge 0: \exists \pi \in SF, x_0^{\pi}=x, x_N^{\pi}\ge
f_N\}$, and $C_*({\mathbb P})=\sup\{x\ge 0: \exists \pi \in SF, x_0^{\pi}=x,
x_N^{\pi}\le f_N\}$. No-arbitrage implies $C_* \le C^*$. Since
\[
\frac{X_N^{\pi}}{B_N} = \frac{X_0^{\pi}}{B_0} +
\sum_{k=1}^N\gamma_k\Delta\left(\frac{S_k}{B_k}\right),
\]
we must have $C_* \le x_* \le x^* \le C^*$. The replicability of
$f_N$ implies $C^*\le C_*$. So, if the market is incomplete but the
contingent claim is still replicable, we still have
$C_*=x_*=x^*=C^*$, i.e. the risk-neutral pricing formula still holds for $t=0$. }
\[
\bar B_t\bar {\mathbb E}\left[\left.\frac{f_T}{\bar B_T}\right|{\cal
F}_t\right]=\widetilde B_t\widetilde {\mathbb E}\left[\left.\frac{f_T}{\widetilde
B_T}\right|{\cal F}_t\right].
\]
That is, the risk-neutral price of $f_T$ is independent of the choice of
num\'{e}raire. See Shiryaev \cite{Shiryaev99}, Chapter VI, \S1b.2.\footnote{The invariance of risk-neutral price gives us a mnemonics to memorize formula (\ref{Radon-Nikodym density for numeraire}): take $f_T=1_A\bar B_T$ with $A\in {\cal F}_T$ and set $t=0$, we have
\[
\bar B_0 \bar {\mathbb P}(A) = \widetilde B_0 \widetilde {\mathbb E} \left[\frac{\bar B_T}{\widetilde B_T}; A\right].
\]
So the Radon-Nikod\'{y}m derivative is $d\bar{\mathbb P}/d\widetilde{\mathbb P} = \frac{\bar B_T/\bar B_0}{\widetilde B_T/\widetilde B_0}$. This leads to formula (9.2.6) in Shreve \cite[page~378]{Shreve04b}. The textbook's chapter summary also provides a good way to memorize: the Radon-Nikod\'{y}m derivative ``is the num\'{e}raire itself, discounted in order to be a martingale and normalized by its initial condition in order to have expected value 1".
}

\medskip

2) The above theoretical results can be applied to market involving
foreign money market account. We consider the following market: a
domestic money market account $M$ ($M_0=1$), a foreign money market
account $M^f$ ($M_0^f=1$), a (vector) asset price process ${\bf S}$ denominated in domestic currency  called
stock. Suppose the domestic vs. foreign currency exchange rate is
$Q$. Note $Q$ is not a traded asset. Denominated by domestic
currency, the traded assets are $(M, M^fQ,{\bf S})$, where $M^fQ$ can be
seen as the price process of one unit foreign currency. Domestic
risk-neutral measure $\widetilde {\mathbb P}$ is such that
$\left(\frac{M^fQ}{M},\frac{{\bf S}}{M}\right)$ is a $\widetilde
{\mathbb P}$-martingale. Denominated by foreign currency, the traded assets
are $\left(M^f, \frac{M}{Q},\frac{\bf S}{Q}\right)$. Foreign
risk-neutral measure $\widetilde {\mathbb P}^f$ is such that
$\left(\frac{M}{QM^f},\frac{\bf S}{QM^f}\right)$ is a $\widetilde
{\mathbb P}^f$-martingale. This is a change of num\'{e}raire in the market
denominated by domestic currency, from $M$ to $M^fQ$. If we assume
the market is arbitrage-free and complete, the foreign risk-neutral
measure is given by
\[
d\widetilde
{\mathbb P}^f=\frac{Q_TM_T^f}{M_T {\mathbb E}\left[\frac{Q_0M_0^f}{M_0}\right]}d\widetilde
{\mathbb P}=\frac{Q_TD_TM_T^f}{Q_0}d\widetilde {\mathbb P}
\]
on ${\cal F}_T$. For a European contingent
claim $f_T$ denominated in domestic currency, its payoff in foreign
currency is $f_T/Q_T$. Therefore its foreign price is $\widetilde
{\mathbb E}^f\left[\left.\frac{D_T^ff_T}{D_t^fQ_T}\right|{\cal F}_t\right]$. Convert this
price into domestic currency, we have $Q_t\widetilde
{\mathbb E}^f\left[\left.\frac{D_T^ff_T}{D_t^fQ_T}\right|{\cal F}_t\right]$. Use the
relation between $\widetilde {\mathbb P}^f$ and $\widetilde {\mathbb P}$ on ${\cal F}_T$
and the Bayes formula, we get
\[
Q_t\widetilde {\mathbb E}^f\left[\left.\frac{D_T^ff_T}{D_t^fQ_T}\right|{\cal
F}_t\right]=\widetilde {\mathbb E}\left[\left.\frac{D_Tf_T}{D_t}\right|{\cal F}_t\right].
\]
The RHS is exactly the price of $f_T$ in domestic market if we
apply risk-neutral pricing.

\medskip

3) {\it Alternative proof of Theorem 9.4.2 (Black-Scholes-Merton option pricing with random interest rate)}. By (9.4.7), $V(t)=B(t,T) \widetilde {\mathbb E}^T[V(T)|{\cal F}(t)]$, where $V(T)=(S(T)-K)^+$ with
\[
S(T)=\mbox{For}_S(T,T)=\mbox{For}_S(t,T)\exp \left\{\sigma[\widetilde W^T(T)-\widetilde W^T(t)]-\frac{1}{2}\sigma^2(T-t)\right\}.
\]
Then the valuation of $\widetilde {\mathbb E}^T[V(T)|{\cal F}(t)]$ is like the Black-Scholes-Merton model for stock options, where interest rate is 0 and the underlying has price $\mbox{For}_S(t,T)$ at time $t$. So its value is
\[
\widetilde {\mathbb E}^T[V(T)|{\cal F}(t)] = \mbox{For}_S(t,T) N(d_+(t)) - K N(d_-(t))
\]
where $d_{\pm}(t)=\frac{1}{\sigma\sqrt{T-t}} \left[\log\frac{\mbox{For}_S(t,T)}{K} \pm \frac{1}{2}\sigma^2(T-t)\right]$. Therefore
\[
V(t) = B(t,T) [\mbox{For}_S(t,T) N(d_+) - K N(d_-)] = S(t) N(d_+(t)) - K B(t,T) N(d_-(t)).
\]

\bigskip

\noindent 9.1. (i)\begin{proof}For any $0\le t\le T$, by Lemma
5.5.2,
\[
E^{(M_2)}\left[\left.\frac{M_1(T)}{M_2(T)}\right|{\cal
F}_t\right]=E\left[\left.\frac{M_2(T)}{M_2(t)}\frac{M_1(T)}{M_2(T)}\right|{\cal
F}_t\right]=\frac{E[M_1(T)|{\cal
F}_t]}{M_2(t)}=\frac{M_1(t)}{M_2(t)}.
\]So $\frac{M_1(t)}{M_2(t)}$ is a martingale under $P^{M_2}$.
\end{proof}

(ii)\begin{proof}Let $M_1(t)=D_tS_t$ and $M_2(t)=D_tN_t/N_0$. Then
$\widetilde P^{(N)}$ as defined in (9.2.6) is $P^{(M_2)}$ as defined
in Remark 9.2.5. Hence $\frac{M_1(t)}{M_2(t)}=\frac{S_t}{N_t}N_0$ is
a martingale under $\widetilde P^{(N)}$, which implies
$S_t^{(N)}=\frac{S_t}{N_t}$ is a martingale under $\widetilde
P^{(N)}$.\end{proof}

\noindent 9.2. (i) \begin{proof}Since $N_t^{-1}=N_0^{-1}e^{-\nu
\widetilde W_t-(r-\frac{1}{2}\nu^2)t}$, we have
\[
d(N_t^{-1})=N_0^{-1}e^{-\nu\widetilde
W_t-(r-\frac{1}{2}\nu^2)t}[-\nu d\widetilde
W_t-(r-\frac{1}{2}\nu^2)dt+\frac{1}{2}\nu^2dt]=N_t^{-1}(-\nu
d\widehat W_t-rdt).
\]
\end{proof}

(ii)\begin{proof}
\[
d\widehat
M_t=M_td\left(\frac{1}{N_t}\right)+\frac{1}{N_t}dM_t+d\left(\frac{1}{N_t}\right)dM_t=\widehat
M_t(-\nu d\widehat W_t-rdt)+r\widehat M_tdt=-\nu \widehat M_t
d\widehat W_t.
\]

{\it Remark:} This can also be obtained directly from Theorem 9.2.2.
\end{proof}

(iii)\begin{proof}
\begin{eqnarray*}
d\widehat
X_t&=&d\left(\frac{X_t}{N_t}\right)=X_td\left(\frac{1}{N_t}\right)+\frac{1}{N_t}dX_t+d\left(\frac{1}{N_t}\right)dX_t\\
&=&(\Delta_tS_t+\Gamma_tM_t)d\left(\frac{1}{N_t}\right)+\frac{1}{N_t}
(\Delta_tdS_t+\Gamma_tdM_t)+d\left(\frac{1}{N_t}\right)(\Delta_tdS_t+\Gamma_tdM_t)\\
&=&\Delta_t\left[S_td\left(\frac{1}{N_t}\right)+\frac{1}{N_t}dS_t+d\left(\frac{1}{N_t}\right)dS_t\right]+\Gamma_t\left[M_td\left(\frac{1}{N_t}\right)+\frac{1}{N_t}dM_t+d\left(\frac{1}{N_t}\right)dM_t\right]\\
&=&\Delta_td\widehat S_t+\Gamma_td\widehat M_t.
\end{eqnarray*}
\end{proof}

\noindent 9.3. To avoid singular cases, we need to assume
$-1<\rho<1$.

(i)\begin{proof} $N_t=N_0e^{\nu \widetilde
W_3(t)+(r-\frac{1}{2}\nu^2)t}$. So
\begin{eqnarray*}
dN_t^{-1}&=&d(N_0^{-1}e^{-\nu \widetilde
W_3(t)-(r-\frac{1}{2}\nu^2)t})\\
&=&N_0^{-1}e^{-\nu\widetilde W_3(t)-(r-\frac{1}{2}\nu^2)t}\left[-\nu
d\widetilde W_3(t)-(r-\frac{1}{2}\nu^2)dt+\frac{1}{2} \nu^2dt\right]\\
&=&N_t^{-1}[-\nu d\widetilde W_3(t)-(r-\nu^2)dt],
\end{eqnarray*}
and
\begin{eqnarray*}
dS_t^{(N)}&=&N_t^{-1}dS_t+S_tdN_t^{-1}+dS_tdN_t^{-1}\\
&=&N_t^{-1}(rS_tdt+\sigma S_td\widetilde W_1(t))+S_tN_t^{-1}[-\nu
d\widetilde W_3(t)-(r-\nu^2)dt]\\
&=&S_t^{(N)}(rdt+\sigma d\widetilde W_1(t))+S_t^{(N)}[-\nu
d\widetilde W_3(t)-(r-\nu^2)dt]-\sigma S_t^{(N)}\rho dt\\
&=&S_t^{(N)}(\nu^2-\sigma\rho)dt+S_t^{(N)}(\sigma d\widetilde
W_1(t)-\nu d\widetilde W_3(t)).
\end{eqnarray*}
Define $\gamma=\sqrt{\sigma^2-2\rho\sigma \nu+\nu^2}$ and
$\widetilde W_4(t)=\frac{\sigma}{\gamma}\widetilde
W_1(t)-\frac{\nu}{\gamma}\widetilde W_3(t)$, then $\widetilde W_4$
is a martingale with quadratic variation
\[
[\widetilde
W_4]_t=\frac{\sigma^2}{\gamma^2}t-2\frac{\sigma\nu}{\gamma^2}\rho
t+\frac{\nu^2}{r^2}t=t.
\]
By L\'{e}vy's Theorem, $\widetilde W_4$ is a BM and therefore,
$S_t^{(N)}$ has volatility
$\gamma=\sqrt{\sigma^2-2\rho\sigma\nu+\nu^2}$.
\end{proof}

(ii)\begin{proof} This problem is the same as Exercise 4.13, we
define $\widetilde W_2(t)=\frac{-\rho}{\sqrt{1-\rho^2}}\widetilde
W_1(t)+\frac{1}{\sqrt{1-\rho^2}}\widetilde W_3(t)$, then $\widetilde
W_2$ is a martingale, with
\[
(d\widetilde
W_2(t))^2=\left(-\frac{\rho}{\sqrt{1-\rho^2}}d\widetilde
W_1(t)+\frac{1}{\sqrt{1-\rho^2}}d\widetilde
W_3(t)\right)^2=\left(\frac{\rho^2}{1-\rho^2}+\frac{1}{1-\rho^2}-\frac{2\rho^2}{1-\rho^2}\right)dt=dt,
\]and $d\widetilde W_2(t)d\widetilde
W_1(t)=-\frac{\rho}{\sqrt{1-\rho^2}}dt+\frac{\rho}{\sqrt{1-\rho^2}}dt=0$.
So $\widetilde W_2$ is a BM independent of $\widetilde W_1$, and
$dN_t=rN_tdt+\nu N_td\widetilde W_3(t)=rN_tdt+\nu N_t[\rho
d\widetilde W_1(t)+\sqrt{1-\rho^2}d\widetilde W_2(t)]$.
\end{proof}

(iii) \begin{proof} Under $\widetilde P$, $(\widetilde
W_1,\widetilde W_2)$ is a two-dimensional BM, and
\begin{eqnarray*}
\begin{cases}
dS_t=rS_tdt+\sigma S_td\widetilde W_1(t)=rS_tdt+S_t(\sigma, 0) \cdot
\left(\begin{matrix} d\widetilde W_1(t)\\ d\widetilde W_2(t)
\end{matrix}\right)\\
dN_t=rN_tdt+\nu N_td\widetilde W_3(t)=rN_tdt+N_t(\nu \rho, \nu
\sqrt{1-\rho^2}) \cdot \left(\begin{matrix} d\widetilde W_1(t)\\
d\widetilde W_2(t)
\end{matrix}\right).
\end{cases}
\end{eqnarray*}
So under $\widetilde P$, the volatility vector for $S$ is $(\sigma,
0)$, and the volatility vector for $N$ is
$(\nu\rho,\nu\sqrt{1-\rho^2})$. By Theorem 9.2.2, under the measure
$\widetilde P^{(N)}$, the volatility vector for $S^{(N)}$ is
$(v_1,v_2)=(\sigma-\nu\rho,-\nu\sqrt{1-\rho^2}$. In particular, the
volatility of $S^{(N)}$ is
\[
\sqrt{v_1^2+v_2^2}=\sqrt{(\sigma-\nu
\rho)^2+(-\nu\sqrt{1-\rho^2})^2}=\sqrt{\sigma^2-2\nu\rho\sigma+\nu^2},
\]consistent with the result of part (i).
\end{proof}

\noindent 9.4. \begin{proof} From (9.3.15), we have
$M_t^fQ_t=M_0^fQ_0e^{\int_0^t\sigma_2(s)d\widetilde
W_3(s)+\int_0^t(R_s-\frac{1}{2}\sigma_2^2(s))ds}$. So
\[
\frac{D_t^f}{Q_t}=D_0^fQ_0^{-1}e^{-\int_0^t\sigma_2(s)d\widetilde
W_3(s)-\int_0^t(R_s-\frac{1}{2}\sigma_2^2(s))ds}\]
 and
\[
d\left(\frac{D_t^f}{Q_t}\right)=\frac{D_t^f}{Q_t}[-\sigma_2(t)d\widetilde
W_3(t)-(R_t-\frac{1}{2}\sigma_2^2(t))dt+\frac{1}{2}\sigma_2^2(t)dt]=\frac{D_t^f}{Q_t}[-\sigma_2(t)d\widetilde
W_3(t)-(R_t-\sigma_2^2(t))dt].
\]To get (9.3.22), we note
\begin{eqnarray*}
d\left(\frac{M_tD_t^f}{Q_t}\right)&=&M_td\left(\frac{D_t^f}{Q_t}\right)+\frac{D_t^f}{Q_t}dM_t+dM_td\left(\frac{D_t^f}{Q_t}\right)\\
&=&\frac{M_tD_t^f}{Q_t}[-\sigma_2(t)d\widetilde
W_3(t)-(R_t-\sigma_2^2(t))dt]+\frac{R_tM_tD_t^f}{Q_t}dt\\
&=&-\frac{M_tD_t^f}{Q_t}(\sigma_2(t)d\widetilde
W_3(t)-\sigma^2_2(t)dt)\\
&=&-\frac{M_tD_t^f}{Q_t}\sigma_2(t)d\widetilde W_3^f(t).
\end{eqnarray*}
To get (9.3.23), we note
\begin{eqnarray*}
d\left(\frac{D_t^fS_t}{Q_t}\right)&=&\frac{D_t^f}{Q_t}dS_t+S_td\left(\frac{D_t^f}{Q_t}\right)+dS_td\left(\frac{D_t^f}{Q_t}\right)\\
&=&\frac{D_t^f}{Q_t}S_t(R_tdt+\sigma_1(t)d\widetilde
W_1(t))+\frac{S_tD_t^f}{Q_t}[-\sigma_2(t)d\widetilde
W_3(t)-(R_t-\sigma_2^2(t))dt]\\
& &+S_t\sigma_1(t)d\widetilde
W_1(t)\frac{D_t^f}{Q_t}(-\sigma_2(t))d\widetilde W_3(t)\\
&=&\frac{D_t^fS_t}{Q_t}[\sigma_1(t)d\widetilde
W_1(t)-\sigma_2(t)d\widetilde
W_3(t)+\sigma_2^2(t)dt-\sigma_1(t)\sigma_2(t)\rho_tdt]\\
&=&\frac{D_t^fS_t}{Q_t}[\sigma_1(t)d\widetilde
W_1^f(t)-\sigma_2d\widetilde W_3^f(t)].
\end{eqnarray*}
\end{proof}


\noindent 9.5. \begin{proof}We combine the solutions of all the
sub-problems into a single solution as follows. The payoff of a
quanto call is $(\frac{S_T}{Q_T}-K)^+$ units of domestic currency at
time $T$. By risk-neutral pricing formula, its price at time $t$ is
$\widetilde E[e^{-r(T-t)}(\frac{S_T}{Q_T}-K)^+|{\cal F}_t]$. So we
need to find the SDE for $\frac{S_t}{Q_t}$ under risk-neutral
measure $\widetilde P$. By formula (9.3.14) and (9.3.16), we have
$S_t=S_0e^{\sigma_1\widetilde W_1(t)+(r-\frac{1}{2}\sigma_1^2)t}$
and
\[
Q_t=Q_0e^{\sigma_2\widetilde
W_3(t)+(r-r^f-\frac{1}{2}\sigma_2^2)t}=Q_0e^{\sigma_2\rho\widetilde
W_1(t)+\sigma_2\sqrt{1-\rho^2}\widetilde
W_2(t)+(r-r^f-\frac{1}{2}\sigma_2^2)t}.
\]So $\frac{S_t}{Q_t}=\frac{S_0}{Q_0}e^{(\sigma_1-\sigma_2\rho)\widetilde W_1(t)-\sigma_2\sqrt{1-\rho^2}\widetilde
W_2(t)+(r^f+\frac{1}{2}\sigma_2^2-\frac{1}{2}\sigma_1^2)t}$. Define
\[
\sigma_4=\sqrt{(\sigma_1-\sigma_2\rho)^2+\sigma_2^2(1-\rho^2)}=\sqrt{\sigma_1^2-2\rho\sigma_1\sigma_2+\sigma_2^2}\;\;\mbox{and}\;\;
\widetilde W_4(t)=\frac{\sigma_1-\sigma_2\rho}{\sigma_4}\widetilde
W_1(t)-\frac{\sigma_2\sqrt{1-\rho^2}}{\sigma_4}\widetilde W_2(t).
\]
Then $\widetilde W_4$ is a martingale with $[\widetilde
W_4]_t=\frac{(\sigma_1-\sigma_2\rho)^2}{\sigma_4^2}t+\frac{\sigma_2(1-\rho^2)}{\sigma_4^2}t+t$.
So $\widetilde W_4$ is a Brownian motion under $\widetilde P$. So if
we set $a=r-r^f+\rho\sigma_1\sigma_2-\sigma_2^2$, we have
\[
\frac{S_t}{Q_t}=\frac{S_0}{Q_0}e^{\sigma_4\widetilde
W_4(t)+(r-a-\frac{1}{2}\sigma_4^2)t}\;\;\mbox{and}\;\;d\left(\frac{S_t}{Q_t}\right)=\frac{S_t}{Q_t}[\sigma_4d\widetilde
W_4(t)+(r-a)dt].
\]Therefore, under $\widetilde P$, $\frac{S_t}{Q_t}$ behaves like
dividend-paying stock and the price of the quanto call option is
like the price of a call option on a dividend-paying stock. Thus
formula (5.5.12) gives us the desired price formula for quanto call
option.
\end{proof}

\noindent 9.6.
(i)\begin{proof}$d_+(t)-d_-(t)=\frac{1}{\sigma\sqrt{T-t}}\sigma^2(T-t)=\sigma\sqrt{T-t}$.
So $d_-(t)=d_+(t)-\sigma\sqrt{T-t}$.\end{proof}

(ii)\begin{proof}$d_+(t)+d_-(t)=\frac{2}{\sigma\sqrt{T-t}}\log\frac{\mbox{For}_S(t,T)}{K}$.
So
\[
d^2_+(t)-d_-^2(t)=(d_+(t)+d_-(t))(d_+(t)-d_-(t))=2\log\frac{\mbox{For}_S(t,T)}{K}.
\]
\end{proof}

(iii)\begin{proof}
\begin{eqnarray*}
\mbox{For}_S(t,T)e^{-d_+^2(t)/2}-Ke^{-d_-^2(t)}&=&e^{-d^2_+(t)/2}[
\mbox{For}_S(t,T)-Ke^{d^2_+(t)/2-d_-^2(t)/2}]\\
&=&e^{-d_+^2(t)/2}[
\mbox{For}_S(t,T)-Ke^{\log\frac{\mbox{For}_S(t,T)}{K}}]\\
&=&0.
\end{eqnarray*}
\end{proof}

(iv)\begin{proof}
\begin{eqnarray*}
&
&dd_+(t)\\
&=&\frac{1}{2}\sqrt{1}{\sigma\sqrt{(T-t)^3}}[\log\frac{\mbox{For}_S(t,T)}{K}+\frac{1}{2}\sigma^2(T-t)]dt+\frac{1}{\sigma\sqrt{T-t}}\left[
\frac{d\mbox{For}_S(t,T)}{\mbox{For}_S(t,T)}-\frac{(d\mbox{For}_S(t,T))^2}{2\mbox{For}_S(t,T)^2}-\frac{1}{2}\sigma
dt\right]\\
&=&\frac{1}{2\sigma\sqrt{(T-t)^3}}\log\frac{\mbox{For}_S(t,T)}{K}dt+\frac{\sigma}{4\sqrt{T-t}}dt+\frac{1}{\sigma\sqrt{T-t}}(\sigma
d\widetilde W^T(t)-\frac{1}{2}\sigma^2dt-\frac{1}{2}\sigma^2dt)\\
&=&\frac{1}{2\sigma(T-t)^{3/2}}\log\frac{\mbox{For}_S(t,T)}{K}dt-\frac{3\sigma}{4\sqrt{T-t}}dt+\frac{d\widetilde
W^T(t)}{\sqrt{T-t}}.
\end{eqnarray*}
\end{proof}

(v)\begin{proof}$dd_-(t)=dd_+(t)-d(\sigma\sqrt{T-t})=dd_+(t)+\frac{\sigma
dt}{2\sqrt{T-t}}$.\end{proof}

(vi)\begin{proof}By (iv) and (v),
$(dd_-(t))^2=(dd_+(t))^2=\frac{dt}{T-t}$.\end{proof}

(vii)\begin{proof}
\begin{eqnarray*}
dN(d_+(t)) &=& N'(d_+(t))dd_+(t)+\frac{1}{2}N''(d_+(t))(dd_+(t))^2\\
&=& \frac{1}{\sqrt{2\pi}}e^{-\frac{d_+^2(t)}{2}}dd_+(t)+\frac{1}{2}\frac{1}{\sqrt{2\pi}}e^{-\frac{d_+^2(t)}{2}}(-d_+(t))\frac{dt}{T-t}.
\end{eqnarray*}
\end{proof}

(viii)\begin{proof}
\begin{eqnarray*}
dN(d_-(t))&=&N'(d_-(t))dd_-(t)+\frac{1}{2}N''(d_-(t))(dd_-(t))^2\\
&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{d_-^2(t)}{2}}\left(dd_+(t)+\frac{\sigma
dt}{2\sqrt{T-t}}\right)+\frac{1}{2}\frac{e^{-\frac{d_-^2(t)}{2}}}{\sqrt{2\pi}}(-d_-(t))\frac{dt}{T-t}\\
&=&\frac{1}{\sqrt{2\pi}}e^{-d^2_-(t)/2}dd_+(t)+\frac{\sigma
e^{-d^2_-(t)/2}}{2\sqrt{2\pi(T-t)}}dt+\frac{e^{-\frac{d^2_-(t)(\sigma\sqrt{T-t}-d_+(t))}{2}}}{2(T-t)\sqrt{2\pi}}dt\\
&=&\frac{1}{\sqrt{2\pi}}e^{-d^2_-(t)/2}dd_+(t)+\frac{\sigma
e^{-d_-^2(t)/2}}{\sqrt{2\pi(T-t)}}dt-\frac{d_+(t)e^{-\frac{d^2_-(t)}{2}}}{2(T-t)\sqrt{2\pi}}dt.
\end{eqnarray*}
\end{proof}

(ix)\begin{proof}
\[
d\mbox{For}_S(t,T)dN(d_+(t))=\sigma \mbox{For}_S(t,T)d\widetilde
W^T(t)\frac{e^{-d^2_+(t)/2}}{\sqrt{2\pi}}\frac{1}{\sqrt{T-t}}d\widehat
W^T(t)=\frac{\sigma
\mbox{For}_S(t,T)e^{-d^2_+(t)/2}}{\sqrt{2\pi(T-t)}}dt.
\]
\end{proof}

(x)\begin{proof}
\begin{eqnarray*}
& &\mbox{For}_S(t,T)dN(d_+(t))+d\mbox{For}_S(t,T)dN(d_+(t))-KdN(d_-(t))\\
&=&\mbox{For}_S(t,T)\left[\frac{1}{\sqrt{2\pi}}e^{-d^2_+(t)/2}dd_+(t)-\frac{d_+(t)}{2(T-t)\sqrt{2\pi}}e^{-d_+^2(t)/2}dt\right]+\frac{\sigma
\mbox{For}_S(t,T)e^{-d^2_+(t)/2}}{\sqrt{2\pi(T-t)}}dt\\
&
&-K\left[\frac{e^{-d^2_-(t)/2}}{\sqrt{2\pi}}dd_+(t)+\frac{\sigma}{\sqrt{2\pi(T-t)}}e^{-d_-^2(t)/2}dt
-\frac{d_+(t)}{2(T-t)\sqrt{2\pi}}e^{-d_-^2(t)/2}dt \right]\\
&=&\left[\frac{\mbox{For}_S(t,T)d_+(t)}{2(T-t)\sqrt{2\pi}}e^{-d_+^2(t)/2}+
\frac{\sigma\mbox{For}_S(t,T)e^{-d^2_+(t)/2}}{\sqrt{2\pi(T-t)}}-\frac{K\sigma
e^{-d^2_-(t)/2}}{\sqrt{2\pi(T-t)}}-\frac{Kd_+(t)}{2(T-t)\sqrt{2\pi}}e^{-d_-^2(t)/2}\right]dt\\
&
&+\frac{1}{\sqrt{2\pi}}\left(\mbox{For}_S(t,T)e^{-d^2_+(t)/2}-Ke^{-d_-^2(t)/2}\right)dd_+(t)\\
&=&0.
\end{eqnarray*} The last ``=" comes from (iii), which implies
$e^{-d_-^2(t)/2}=\frac{\mbox{For}_S(t,T)}{K}e^{-d_+^2(t)/2}$.
\end{proof}

\medskip

\section{Term-Structure Models}

$\bigstar$ {\bf Comments}:

\medskip

1) {\it Computation of $e^{\Lambda t}$ (Lemma 10.2.3)}. For a systematic treatment of the computation of matrix exponential function $e^{\Lambda t}$, see 丁同仁等 \cite[page~175]{DL91} or Arnold et al. \cite[page~221]{AC92}. For the sake of Lemma 10.2.3, a direct computation to find (10.2.35) goes as follows.

i) The case of $\lambda_1=\lambda_2$. In this case, set $\Gamma_1 = \left[\begin{matrix}\lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix}\right]$ and $\Gamma_2 = \left[\begin{matrix}0 & 0 \\ \lambda_{21} & 0 \end{matrix}\right]$. Then $\Lambda=\Gamma_1+\Gamma_2$, $\Gamma_2^n = 0_{2\times2}$ for $n\ge 2$, and $\Gamma_1^n = \left[\begin{matrix}\lambda_1^n & 0 \\ 0 & \lambda_2^n \end{matrix}\right]$. Note $\Gamma_1$ and $\Gamma_2$ commute, we therefore have
\[
e^{\Lambda t} = e^{\Gamma_1 t} \cdot e^{\Gamma_2 t}
= \left[\begin{matrix}e^{\lambda_1t} & 0 \\ 0 & e^{\lambda_2t} \end{matrix}\right]\left(I_{2\times2} + \left[\begin{matrix}0 & 0 \\ \lambda_{21}t  & 0 \end{matrix}\right]\right) = \left[\begin{matrix}e^{\lambda_1t} & 0 \\ \lambda_{21} t e^{\lambda_2t} & e^{\lambda_2t} \end{matrix}\right]
= \left[\begin{matrix}e^{\lambda_1t} & 0 \\ \lambda_{21} t e^{\lambda_1t} & e^{\lambda_2t} \end{matrix}\right].
\]

ii) The case of $\lambda_1\ne\lambda_2$. In this case, $\Lambda$ has two distinct eigenvalues $\lambda_1$ and $\lambda_2$. So it can be diagonalized, that is, we can find a matrix $P$ such that
\[
P^{-1}\Lambda P = J = \left[\begin{matrix}\lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix}\right].
\]
Solving the matrix equation $P\Lambda = PJ$, we have
\[
P = \left[\begin{matrix}1 & 0 \\ \frac{\lambda_{21}}{\lambda_1-\lambda_2} & 1 \end{matrix}\right],
\]
and consequently
\[
P^{-1} = \left[\begin{matrix}1 & 0 \\ -\frac{\lambda_{21}}{\lambda_1-\lambda_2} & 1 \end{matrix}\right].
\]
Therefore
\[
e^{\Lambda t} = \sum_{n=0}^{\infty}\frac{PJ^nP^{-1}}{n!}t^n = P \cdot \left(\sum_{n=0}^{\infty}\frac{J^n}{n!}t^n \right) \cdot P^{-1} = P \left[\begin{matrix}e^{\lambda_1t} & 0 \\ 0 & e^{\lambda_2t} \end{matrix}\right] P^{-1} = \left[\begin{matrix}e^{\lambda_1t} & 0 \\ \frac{\lambda_{21}}{\lambda_1-\lambda_2}\left(e^{\lambda_1t}-e^{\lambda_2t}\right) & e^{\lambda_2t} \end{matrix}\right].
\]

\medskip

2) Intuition of Theorem 10.4.1 (Price of backset LIBOR). The intuition here is the same as that of a discrete-time model: the payoff $\delta L(T,T) =  (1+\delta L(T,T)) - 1$ at time $T+\delta$ is equivalent to an investment of 1 at time $T$ (which gives the payoff $1+\delta L(T,T)$ at time $T+\delta$), subtracting a payoff of 1 at time $T+\delta$. The first part has time $t$ price $B(t,T)$ while the second part has time $t$ price $B(t,T+\delta)$.

{\it Pricing under $(T+\delta)$-forward measure}: Using the $(T+\delta)$-forward measure, the time-$t$ no-arbitrage price of a payoff $\delta L(T,T)$ at time $(T+\delta)$ is
\[
B(t,T+\delta) \widetilde {\mathbb E}^{T+\delta}[\delta L(T,T)] = B(t,T+\delta) \cdot \delta L(t,T),
\]
where we have used the observation that the forward rate process $L(t,T) = \frac{1}{\delta}\left(\frac{B(t,T)}{B(t,T+\delta)} - 1\right) \; (0 \le t \le T)$ is a martingale under the $(T+\delta)$-forward measure.

\bigskip

\noindent $\blacktriangleright$ {\bf Exercise 10.1 (Statistics in the two-factor Vasicek model).} According to Example 4.7.3, $Y_1(t)$ and $Y_2(t)$ in (10.2.43)-(10.2.46) are Gaussian processes.

\smallskip

\noindent (i) Show that
\[
\widetilde {\mathbb E} Y_1(t) = e^{-\lambda_1t}Y_1(0), \tag{10.7.1}
\]
that when $\lambda_1\ne \lambda_2$, then
\[
\widetilde {\mathbb E}Y_2(t) = \frac{\lambda_{21}}{\lambda_1-\lambda_2} \left(e^{-\lambda_1 t}-e^{-\lambda_2 t}\right) Y_1(0) + e^{-\lambda_2 t}Y_2(0), \tag{10.7.2}
\]
and when $\lambda_1=\lambda_2$, then
\[
\widetilde {\mathbb E}Y_2(t) = -\lambda_{21} t e^{-\lambda_1t}Y_1(0) + e^{-\lambda_1t}Y_2(0). \tag{10.7.3}
\]

We can write
\[
Y_1(t) - \widetilde {\mathbb E}Y_1(t) = e^{-\lambda_1 t} I_1(t),
\]
when $\lambda_1 \ne \lambda_2$,
\[
Y_2(t) - \widetilde{\mathbb E}Y_2(t) = \frac{\lambda_{21}}{\lambda_1 - \lambda_2} \left(e^{-\lambda_1t}I_1(t)-e^{-\lambda_2t}I_2(t)\right) - e^{-\lambda_2t} I_3(t),
\]
and when $\lambda_1=\lambda_2$,
\[
Y_2(t) - \widetilde{\mathbb E}Y_2(t) = -\lambda_{21} t e^{-\lambda_1 t}I_1(t) + \lambda_{21}e^{-\lambda_1t}I_4(t) + e^{-\lambda_1 t} I_3(t),
\]
where the It\^{o} integrals
\begin{eqnarray*}
I_1(t) = \int_0^t e^{\lambda_1 u} d\widetilde W_1(u), & & I_2(t) = \int_0^t e^{\lambda_2u}d\widetilde W_1(u), \\
I_3(t) = \int_0^t e^{\lambda_2 u} d\widetilde W_2(u), & & I_4(t) = \int_0^t u e^{\lambda_1 u} d\widetilde W_1(u),
\end{eqnarray*}
all have expectation zero under the risk-neutral measure $\widetilde {\mathbb P}$. Consequently, we can determine the variances of $Y_1(t)$ and $Y_2(t)$ and the covariance of $Y_1(t)$ and $Y_2(t)$ under the risk-neutral measure from the variances and covariances of $I_j(t)$ and $I_k(t)$. For example, if $\lambda_1 = \lambda_2$, then
\begin{eqnarray*}
\mbox{Var}(Y_1(t)) &=& e^{-2\lambda_1 t}\widetilde {\mathbb E} I_1^2(t), \\
\mbox{Var}(Y_2(t)) &=& \lambda_{21}^2 t^2 e^{-2\lambda_1t} \widetilde {\mathbb E}I_1^2(t) + \lambda_{21}^2 e^{-2\lambda_1t}\widetilde {\mathbb E}I_4^2(t) + e^{-2\lambda_1t}\widetilde {\mathbb E}I_3^2(t)\\
& & -2\lambda_{21}^2 t e^{-2\lambda_1t} \widetilde{\mathbb E}[I_1(t)I_4(t)] - 2\lambda_{21} t e^{-2\lambda_1t}\widetilde{\mathbb E}[I_1(t)I_3(t)]\\
& & + 2\lambda_{21} e^{-2\lambda_1t}\widetilde{\mathbb E}[I_4(t)I_3(t)], \\
\mbox{Cov}(Y_1(t), Y_2(t)) &=& -\lambda_{21}te^{-2\lambda_1t}\widetilde{\mathbb E}I_1^2(t) + \lambda_{21}e^{-2\lambda_1t}\widetilde {\mathbb E}[I_1(t)I_4(t)] + e^{-2\lambda_1t}\widetilde{\mathbb E}[I_1(t)I_3(t)],
\end{eqnarray*}
where the variances and covariance above are under the risk-neutral measure $\widetilde {\mathbb P}$.

\begin{proof} Using the notation $I_1(t)$,
$I_2(t)$, $I_3(t)$ and $I_4(t)$ introduced in the problem, we can
write $Y_1(t)$ and $Y_2(t)$ as
\[
Y_1(t)=e^{-\lambda_1t}Y_1(0)+e^{-\lambda_1t}I_1(t)
\]
and
\begin{eqnarray*}
& &Y_2(t) \\
&=&
\begin{cases}
\frac{\lambda_{21}}{\lambda_1-\lambda_2}(e^{-\lambda_1t}-e^{-\lambda_2t})Y_1(0)+e^{-\lambda_2t}Y_2(0)+\frac{\lambda_{21}}{\lambda_1-\lambda_2}\left[e^{-\lambda_1t}I_1(t)-e^{-\lambda_2t}I_2(t)\right]-e^{-\lambda_2t}I_3(t),&
\mbox{$\lambda_1\ne\lambda_2$;}\\
-\lambda_{21}te^{-\lambda_1t}Y_1(0)+e^{-\lambda_1t}Y_2(0)-\lambda_{21}\left[te^{-\lambda_1t}I_1(t)-e^{-\lambda_1t}I_4(t)\right]+e^{-\lambda_1t}I_3(t),&\mbox{$\lambda_1=\lambda_2$.}
\end{cases}
\end{eqnarray*}
Since all the $I_k(t)$'s $(k=1,\cdots,4)$ are normally distributed
with zero mean, we can conclude
\[
\widetilde {\mathbb E}[Y_1(t)]=e^{-\lambda_1t}Y_1(0)
\]
and
\begin{eqnarray*}
\widetilde {\mathbb E}[Y_2(t)]=
\begin{cases}
\frac{\lambda_{21}}{\lambda_1-\lambda_2}(e^{-\lambda_1t}-e^{-\lambda_2t})Y_1(0)+e^{-\lambda_2t}Y_2(0),&
\mbox{if $\lambda_1\ne\lambda_2$;}\\
-\lambda_{21}te^{-\lambda_1t}Y_1(0)+e^{-\lambda_1t}Y_2(0),&\mbox{if
$\lambda_1=\lambda_2$.}
\end{cases}
\end{eqnarray*}
\end{proof}

\noindent (ii) Compute the five terms
\[
\widetilde{\mathbb E}I_1^2(t), \; \widetilde{\mathbb E}[I_1(t)I_2(t)], \; \widetilde{\mathbb E}[I_1(t)I_3(t)],
\; \widetilde{\mathbb E}[I_1(t)I_4(t)], \; \widetilde{\mathbb E}[I_4^2(t)].
\]
The five other terms, which you are not being asked to compute, are
\begin{eqnarray*}
{\mathbb E}I_2^2(t) &=& \frac{1}{2\lambda_2} \left(e^{2\lambda_2t}-1\right), \\
{\mathbb E}[I_2(t)I_3(t)] &=& 0, \\
{\mathbb E}[I_2(t)I_4(t)] &=& \frac{t}{\lambda_1+\lambda_2}e^{(\lambda_1+\lambda_2)t} + \frac{1}{(\lambda_1+\lambda_2)^2}\left(1 - e^{(\lambda_1+\lambda_2)t}\right), \\
{\mathbb E}I_3^2(t) &=& \frac{1}{\lambda_2} \left(e^{2\lambda_2 t}-1\right), \\
{\mathbb E}[I_3(t)I_4(t)] &=& 0.
\end{eqnarray*}

\begin{solution} The calculation relies on the following fact: if
$X_t$ and $Y_t$ are both martingales, then $X_tY_t-[X,Y]_t$ is also
a martingale. In particular, $\widetilde {\mathbb E}[X_tY_t]=\widetilde
{\mathbb E}\{[X,Y]_t\}$. Thus
\[
\widetilde
{\mathbb E}[I_1^2(t)]=\int_0^te^{2\lambda_1u}du=\frac{e^{2\lambda_1t}-1}{2\lambda_1},
\; \widetilde
{\mathbb E}[I_1(t)I_2(t)]=\int_0^te^{(\lambda_1+\lambda_2)u}du=\frac{e^{(\lambda_1+\lambda_2)t}-1}{\lambda_1+\lambda_2},
\]
\[
\widetilde {\mathbb E}[I_1(t)I_3(t)]=0,\;\widetilde
{\mathbb E}[I_1(t)I_4(t)]=\int_0^tue^{2\lambda_1u}du=\frac{1}{2\lambda_1}\left[te^{2\lambda_1t}-\frac{e^{2\lambda_1t}-1}{2\lambda_1}\right]
\]and
\begin{eqnarray*}
\widetilde
{\mathbb E}[I_4^2(t)]=\int_0^tu^2e^{2\lambda_1u}du=\frac{t^2e^{2\lambda_1t}}{2\lambda_1}-\frac{te^{2\lambda_1t}}{2\lambda_1^2}+\frac{e^{2\lambda_1t}-1}{4\lambda_1^3}.
\end{eqnarray*}
\end{solution}

\noindent (iii) Some derivative securities involve {\it time spread} (i.e., they depend on the interest rate at two different times). In such cases, we are interested in the joint statistics of the factor processes at different times. These are still jointly normal and depend on the statistics of the It\^{o} integral $I_j$ at different times. Compute $\widetilde {\mathbb E}[I_1(s)I_2(t)]$, where $0\le s < t$. (Hint: Fix $s\ge 0$ and define
\[
J_1(t) = \int_0^t e^{\lambda_1 u} 1_{\{u\le s\}} d\widetilde W_1(u),
\]
where $1_{\{u\le s\}}$ is the function of $u$ that is 1 if $u\le s$ and 0 if $u>s$. Note that $J_1(t)=I_1(s)$ when $t\ge s$.)

\begin{solution} Following the hint, we have
\begin{eqnarray*}
\widetilde {\mathbb E}[I_1(s)I_2(t)]=\widetilde
{\mathbb E}[J_1(t)I_2(t)]=\int_0^te^{(\lambda_1+\lambda_2)u}1_{\{u\le
s\}}du=\frac{e^{(\lambda_1+\lambda_2)s}-1}{\lambda_1+\lambda_2}.
\end{eqnarray*}
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 10.2 (Ordinary differential equations for the mixed affine-yield model).} In the mixed model of Subsection 10.2.3, as in the two-factor Cox-Ingersoll-Ross model, zero-coupon bond prices have the affine-yield form
\[
f(t,y_1,y_2)=e^{-y_1C_1(T-t)-y_2C_2(T-t)-A(T-t)},
\]
where $C_1(0)=C_2(0)=A(0)=0$.

\smallskip

\noindent (i) Find the partial differential equation satisfied by $f(t,y_1,y_2)$.

\begin{solution}
Assume $B(t, T)= \widetilde {\mathbb E}\left[\left.e^{-\int_t^TR_sds}\right|{\cal F}_t\right]=f(t,Y_1(t),Y_2(t))$. Then
\[
d(D(t)B(t,T))=D(t)[-R(t)f(t,Y_1(t),Y_2(t))dt+ df(t,Y_1(t),Y_2(t))].
\]
By It\^{o}'s formula,
\begin{eqnarray*}
df(t,Y_1(t),Y_2(t))&=&  \left[f_t(t,Y_1(t),Y_2(t))+f_{y_1}(t,Y_1(t),Y_2(t))(\mu-\lambda_1Y_1(t))+f_{y_2}(t,Y_1(t),Y_2(t))(-\lambda_2)Y_2(t) \right.\\
& & + f_{y_1y_2}(t,Y_1(t),Y_2(t))\sigma_{21}Y_1(t)+\frac{1}{2}f_{y_1y_1}(t,Y_1(t),Y_2(t))Y_1(t)\\
& & \left. + \frac{1}{2}f_{y_2y_2}(t,Y_1(t),Y_2(t))(\sigma_{21}^2Y_1(t)+\alpha+\beta Y_1(t))\right]dt+\mbox{martingale part}.\\
\end{eqnarray*}
Since $D(t)B(t,T)$ is a martingale, we must have
\begin{eqnarray*}
& & \left[-(\delta_0+\delta_1y_1+\delta_2y_2)+\frac{\partial}{\partial t}+(\mu-\lambda_1y_1)\frac{\partial}{\partial y_1}-\lambda_2y_2\frac{\partial}{\partial y_2} \right]f \\
& & +\left[\frac{1}{2}\left(2\sigma_{21}y_1\frac{\partial^2}{\partial y_1\partial y_2}+y_1\frac{\partial^2}{\partial y_1^2}+(\sigma_{21}^2y_1+\alpha+\beta y_1)\frac{\partial^2}{\partial y_2^2}\right)\right]f \\
&=& 0.
\end{eqnarray*}
\end{solution}

\noindent (ii) Show that $C_1$, $C_2$, and $A$ satisfy the system of ordinary differential equations\footnote{The textbook has a typo in formula (10.7.4): the coefficient of $C_2^2$ should be $- \frac{1}{2}(\sigma^2_{21}+\beta)$ instead of $-(1+\beta)$. See \href{http://www.math.cmu.edu/~shreve/}{http://www.math.cmu.edu/$\sim$shreve/} for details.}
\[
C_1' = -\lambda_1 C_1 - \frac{1}{2}C_1^2 - \sigma_{21}C_1 C_2 - \frac{1}{2}(\sigma^2_{21}+\beta) C_2^2 + \delta_1, \tag{10.7.4}
\]
\[
C_2' = -\lambda_2C_2 + \delta_2, \tag{10.7.5}
\]
\[
A'  = \mu C_1 - \frac{1}{2}\alpha C_2^2 + \delta_0.  \tag{10.7.6}
\]
\begin{proof}
If we suppose  $f(t,y_1,y_2)=e^{-y_1C_1(T-t)-y_2C_2(T-t)-A(T-t)}$, then
\begin{eqnarray*}
\frac{\partial f}{\partial t}  &=& [y_1C_1'(T-t)+y_2C_2'(T-t)+A'(T-t)]f \\
\frac{\partial f}{\partial y_1}  &=& -C_1(T-t)f \\
\frac{\partial f}{\partial y_2} &=& -C_2(T-t)f \\
\frac{\partial^2 f}{\partial y_1\partial y_2} &=& C_1(T-t)C_2(T-t)f \\
\frac{\partial^2 f}{\partial y_1^2} &=& C_1^2(T-t)f \\
\frac{\partial^2f}{\partial y_2^2} &=& C_2^2(T-t)f.
\end{eqnarray*}
So the PDE in part (i) becomes
\begin{eqnarray*}
& & -(\delta_0+\delta_1y_1+\delta_2y_2)+y_1C_1'+y_2C_2'+A'-(\mu-\lambda_1y_1)C_1+
\lambda_2y_2C_2 \\
& & +\frac{1}{2}\left[2\sigma_{21}y_1C_1C_2+y_1C_1^2
+(\sigma^2_{21}y_1+\alpha+\beta y_1)C_2^2\right] \\
&=& 0.
\end{eqnarray*}
Sorting out the LHS according to the independent variables $y_1$ and $y_2$, we get
\begin{eqnarray*}
\begin{cases}
-\delta_1+C_1'+\lambda_1C_1+\sigma_{21}C_1C_2+\frac{1}{2}C_1^2+
\frac{1}{2}(\sigma^2_{21}+\beta)C_2^2=0 \\
-\delta_2+C_2'+\lambda_2C_2=0\\
-\delta_0+A'-\mu C_1 +\frac{1}{2}\alpha C_2^2=0.
\end{cases}
\end{eqnarray*}
In other words, we can obtain the ODEs for $C_1, C_2$ and $A$ as follows
\begin{eqnarray*}
\begin{cases}
C_1'=-\lambda_1C_1-\sigma_{21}C_1C_2-\frac{1}{2}C_1^2-\frac{1}{2}(\sigma^2_{21}+\beta)C_2^2+\delta_1 \\
C_2'=-\lambda_2C_2 +\delta_2 \\
A'= \mu C_1 -\frac{1}{2}\alpha C_2^2+\delta_0.
\end{cases}
\end{eqnarray*}

\end{proof}

\noindent $\blacktriangleright$ {\bf Exericse 10.3 (Calibration of the two-factor Vasicek model).} Consider the canonical two-factor Vasicek model (10.2.4), (10.2.5), but replace the interest rate equation (10.2.6) by
\[
R(t) = \delta_0(t) + \delta_1 Y_1(t) + \delta_2 Y_2(t), \tag{10.7.7}
\]
where $\delta_1$ and $\delta_2$ are constant but $\delta_0(t)$ is a nonrandom function of time. Assume that for each $T$ there is a zero-coupon bond maturing at time $T$. The price of this bond at time $t\in [0, T]$ is
\[
B(t,T) = \widetilde {\mathbb E} \left[\left.e^{-\int_t^T R(u)du}\right|{\cal F}(t)\right].
\]
Because the pair of processes $(Y_1(t), Y_2(t))$ is Markov, there must exist some function $f(t,T,y_1,y_2)$ such that $B(t,T)=f(t,T,Y_1(t),Y_2(t))$. (We indicate the dependence of $f$ on the maturity $T$ because, unlike in Subsection 10.2.1, here we shall consider more than one value of $T$.)

\smallskip

\noindent (i) The function $f(t,T,y_1,y_2)$ is of the affine-yield form
\[
f(t,T,y_1,y_2)=e^{-y_1C_1(t,T)-y_2C_2(t,T)-A(t,T)}. \tag{10.7.8}
\]
Holding $T$ fixed, derive a system of ordinary differential equations for $\frac{d}{dt}C_1(t,T)$, $\frac{d}{dt}C_2(t,T)$, and $\frac{d}{dt}A(t,T)$.

\begin{solution}
We have $d(D_tB(t,T))=D_t[-R_tf(t,T,Y_1(t),Y_2(t))dt+df(t,T,Y_1(t),Y_2(t))]$ and \begin{eqnarray*}
& & df(t,T,Y_1(t),Y_2(t))\\
&=&[f_t(t,T,Y_1(t),Y_2(t))+f_{y_1}(t,T,Y_1(t),Y_2(t))(-\lambda_1Y_1(t))+f_{y_2}(t,T,Y_1(t),Y_2(t))(-\lambda_{21}Y_1(t)-\lambda_2Y_2(t))\\
& &+\frac{1}{2}f_{y_1y_1}(t,T,Y_1(t),Y_2(t))+\frac{1}{2}f_{y_2y_2}(t,T,Y_1(t),Y_2(t))]dt + \mbox{martingale part}.
\end{eqnarray*}
Since $D_tB(t,T)$ is a martingale under risk-neutral measure, we have the following PDE:
\[
\left[-(\delta_0(t)+\delta_1y_1+\delta_2y_2)+\frac{\partial}{\partial t}-\lambda_1y_1\frac{\partial}{\partial y_1}-(\lambda_{21}y_1+\lambda_2y_2)\frac{\partial}{\partial y_2}+\frac{1}{2}\frac{\partial^2}{\partial y_1^2}+\frac{1}{2}\frac{\partial}{\partial y_2^2}\right]f(t,T,y_1,y_2)=0.
\]
Suppose $f(t,T,y_1,y_2)=e^{-y_1C_1(t,T)-y_2C_2(t,T)-A(t,T)}$, then
\begin{eqnarray*}
\begin{cases}
f_t(t,T,y_1,y_2) =\left[-y_1\frac{d}{dt}C_1(t,T)-y_2\frac{d}{dt}C_2(t,T)-\frac{d}{dt}A(t,T)\right]f(t,T,y_1,y_2),\\
f_{y_1}(t,T,y_1,y_2)=-C_1(t,T)f(t,T,y_1,y_2),\\
f_{y_2}(t,T,y_1,y_2)=-C_2(t,T)f(t,T,y_1,y_2),\\
f_{y_1y_2}(t,T,y_1,y_2)=C_1(t,T)C_2(t,T)f(t,T,y_1,y_2),\\
f_{y_1y_1}(t,T,y_1,y_2)=C_1^2(t,T)f(t,T,y_1,y_2),\\
f_{y_2y_2}(t,T,y_1,y_2)=C_2^2(t,T)f(t,T,y_1,y_2).
\end{cases}
\end{eqnarray*}
So the PDE becomes
\begin{eqnarray*}
& &-(\delta_0(t)+\delta_1y_1+\delta_2y_2)+\left(-y_1\frac{d}{dt}C_1(t,T)-y_2\frac{d}{dt}C_2(t,T)-\frac{d}{dt}A(t,T)\right)
+\lambda_1y_1C_1(t,T)\\
& &+(\lambda_{21}y_1 + \lambda_2y_2)C_2(t,T)+\frac{1}{2}C_1^2(t,T)+\frac{1}{2}C_2^2(t,T)=0.
\end{eqnarray*}
Sorting out the terms according to independent variables $y_1$ and $y_2$, we get
\begin{eqnarray*}
\begin{cases}
-\delta_0(t)-\frac{d}{dt}A(t,T)+\frac{1}{2}C_1^2(t,T)+\frac{1}{2}C_2^2(t,T)=0\\
-\delta_1-\frac{d}{dt}C_1(t,T)+\lambda_1C_1(t,T)+\lambda_{21}C_2(t,T)=0\\
-\delta_2-\frac{d}{dt}C_2(t,T)+\lambda_2C_2(t,T)=0.
\end{cases}
\end{eqnarray*}
That is
\begin{eqnarray*}
\begin{cases}
\frac{d}{dt}C_1(t,T)=\lambda_1C_1(t,T)+\lambda_{21}C_2(t,T)-\delta_1\\
\frac{d}{dt}C_2(t,T)=\lambda_2C_2(t,T)-\delta_2\\
\frac{d}{dt}A(t,T)=\frac{1}{2}C_1^2(t,T)+\frac{1}{2}C_2^2(t,T)-\delta_0(t).
\end{cases}
\end{eqnarray*}
\end{solution}

\noindent (ii) Using the terminal conditions $C_1(T,T)=C_2(T,T)=0$, solve the equations in (i) for $C_1(t,T)$ and $C_2(t,T)$. (As in Subsection 10.2.1, the functions $C_1$ and $C_2$ depend on $t$ and $T$ only through the difference $\tau=T-t$; however, the function $A$ discussed in part (iii) below depends on $t$ and $T$ separately.)

\begin{solution}
For $C_2$, we note $\frac{d}{dt}[e^{-\lambda_2t}C_2(t,T)]=-e^{-\lambda_2t}\delta_2$ from the ODE in (i). Integrate from $t$ to $T$, we have $0-e^{-\lambda_2t}C_2(t,T)=-\delta_2\int_t^Te^{-\lambda_2s}ds=\frac{\delta_2}{\lambda_2}(e^{-\lambda_2T}-e^{-\lambda_2t})$. So $C_2(t,T)=\frac{\delta_2}{\lambda_2} \left(1-e^{-\lambda_2(T-t)}\right)$. For $C_1$, we note
\[
\frac{d}{dt}(e^{-\lambda_1t}C_1(t,T))=(\lambda_{21}C_2(t,T)-\delta_1)e^{-\lambda_1t}=\frac{\lambda_{21}\delta_2}{\lambda_2}
(e^{-\lambda_1t}-e^{-\lambda_2T+(\lambda_2-\lambda_1)t})-\delta_1e^{-\lambda_1t}.
\]
Integrate from $t$ to $T$, we get
\begin{eqnarray*}
& & -e^{-\lambda_1t}C_1(t,T) \\
&=& \begin{cases}
-\frac{\lambda_{21}\delta_2}{\lambda_2\lambda_1}(e^{-\lambda_1T}-e^{-\lambda_1t})-\frac{\lambda_{21}\delta_2}{\lambda_2}e^{-\lambda_2T}
\frac{e^{(\lambda_2-\lambda_1)T}-e^{(\lambda_2-\lambda_1)t}}{\lambda_2-\lambda_1} + \frac{\delta_1}{\lambda_1}(e^{-\lambda_1T}-e^{-\lambda_1T}) & \mbox{if $\lambda_1 \ne \lambda_2$} \\
-\frac{\lambda_{21}\delta_2}{\lambda_2\lambda_1}(e^{-\lambda_1T}-e^{-\lambda_1t})-\frac{\lambda_{21}\delta_2}{\lambda_2}e^{-\lambda_2T}
(T-t) + \frac{\delta_1}{\lambda_1}(e^{-\lambda_1T}-e^{-\lambda_1T}) & \mbox{if $\lambda_1 = \lambda_2$}.
\end{cases}
\end{eqnarray*}
So
\begin{eqnarray*}
C_1(t,T)=\begin{cases}
\frac{\lambda_{21}\delta_2}{\lambda_2\lambda_1}(e^{-\lambda_1(T-t)}-1)+\frac{\lambda_{21}\delta_2}{\lambda_2}\frac{e^{-\lambda_1(T-t)}-e^{-\lambda_2(T-t)}}{\lambda_2-\lambda_1}-\frac{\delta_1}{\lambda_1}(e^{-\lambda_1(T-t)}-1) & \mbox{if $\lambda_1 \ne \lambda_2$}\\
\frac{\lambda_{21}\delta_2}{\lambda_2\lambda_1}(e^{-\lambda_1(T-t)}-1)+\frac{\lambda_{21}\delta_2}{\lambda_2}e^{-\lambda_2T+\lambda_1t}(T-t)-\frac{\delta_1}{\lambda_1}(e^{-\lambda_1(T-t)}-1) & \mbox{if $\lambda_1 = \lambda_2$}.
\end{cases}.
\end{eqnarray*}
\end{solution}

\noindent (iii) Using the terminal condition $A(T,T)=0$, write a formula for $A(t,T)$ as an integral involving $C_1(u,T)$, $C_2(u,T)$, and $\delta_0(u)$. You do not need to evaluate this integral.

\begin{solution}
From the ODE $\frac{d}{dt}A(t,T)=\frac{1}{2}(C_1^2(t,T)+C_2^2(t,T))-\delta_0(t)$, we get
\[
A(t,T)=\int_t^T\left[\delta_0(s)-\frac{1}{2}(C_1^2(s,T)+C_2^2(s,T))\right]ds.
\]
\end{solution}

\noindent (iv) Assume that the model parameters $\lambda_1>0$ $\lambda_2 > 0$, $\lambda_{21}$, $\delta_1$, and $\delta_2$ and the initial conditions $Y_1(0)$ and $Y_2(0)$ are given. We wish to choose a {\it function} $\delta_0$ so that the zero-coupon bond prices given by the model match the bond prices given by the market at the initial time zero. In other words, we want to choose a function $\delta(T)$, $T\ge 0$, so that
\[
f(0, T, Y_1(0), Y_2(0)) = B(0, T), \; T\ge 0.
\]
In this part of the exercise, we regard both $t$ and $T$ as variables and uses the notation $\frac{\partial}{\partial t}$ to indicate the derivative with respect to $t$ when $T$ is held fixed and the notation $\frac{\partial }{\partial T}$ to indicate the derivative with respect to $T$ when $t$ is held fixed. Give a formula for $\delta_0(T)$ in terms of $\frac{\partial}{\partial T}\log B(0,T)$ and the model parameters. (Hint: Compute $\frac{\partial}{\partial T}A(0,T)$ in two ways, using (10.7.8) and also using the formula obtained in (iii). Because $C_i(t,T)$ depends only on $t$ and $T$ through $\tau=T-t$, there are functions $\overline{C}_i(\tau)$ such that $\overline{C}_i(\tau)=\overline{C}_i(T-t)=C_i(t,T)$, $i=1,2$. Then
\[
\frac{\partial}{\partial t}C_i(t,T) = - \overline{C}_{i}'(\tau), \frac{\partial }{\partial T} C_i(t,T)=\overline{C}_i'(\tau),
\]
where $'$ denotes differentiation with respect to $\tau$. This shows that
\[
\frac{\partial}{\partial T}C_i(t,T) = -\frac{\partial}{\partial t}C_i(t,T), \; i=1,2, \tag{10.7.9}
\]
a fact that you will need.)

\begin{solution}
We want to find $\delta_0$ so that $f(0,T,Y_1(0),Y_2(0))=e^{-Y_1(0)C_1(0,T)-Y_2(0)C_2(0,T)-A(0,T)}=B(0,T)$ for
all $T>0$. Take logarithm on both sides and plug in the expression of $A(t,T)$, we get
\[
\log B(0,T)=-Y_1(0)C_1(0,T)-Y_2(0)C_2(0,T)+\int_0^T\left[\frac{1}{2}(C_1^2(s,T)+C_2^2(s,T))-\delta_0(s)\right]ds.
\]
Taking derivative w.r.t. $T$, we have
\[
\frac{\partial}{\partial T}\log B(0,T)=-Y_1(0)\frac{\partial}{\partial T}C_1(0,T)-Y_2(0)\frac{\partial}{\partial T}
C_2(0,T)+\frac{1}{2}C_1^2(T,T)+\frac{1}{2}C_2^2(T,T)-\delta_0(T).
\]
Therefore
\begin{eqnarray*}
\delta_0(T) &=& -Y_1(0)\frac{\partial}{\partial T}C_1(0,T)-Y_2(0)\frac{\partial}{\partial T}C_2(0,T)-
\frac{\partial}{\partial T}\log B(0,T) \\
&=& \begin{cases}
-Y_1(0)\left[\delta_1e^{-\lambda_1T}-\frac{\lambda_{21}\delta_2}{\lambda_2}e^{-\lambda_2T}\right]-Y_2(0)\delta_2e^{-\lambda_2T}-\frac{\partial}{\partial T}\log B(0,T) & \mbox{if $\lambda_1 \ne \lambda_2$}\\
-Y_1(0)\left[\delta_1e^{-\lambda_1T}-\lambda_{21}\delta_2e^{-\lambda_2T}T\right]-Y_2(0)\delta_2e^{-\lambda_2T}-\frac{\partial}{\partial T}\log B(0,T) & \mbox{if $\lambda_1 = \lambda_2$}.
\end{cases}
\end{eqnarray*}
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 10.4.} Hull and White [89] propose the two-factor model
\[
dU(t) = -\lambda_1 U(t) dt + \sigma_1 d\widetilde B_2(t),  \tag{10.7.10}
\]
\[
dR(t) = [\theta(t)+U(t)-\lambda_2R(t)]dt + \sigma_2 d\widetilde B_1(t), \tag{10.7.11}
\]
where $\lambda_1$, $\lambda_2$, $\sigma_1$, and $\sigma_2$ are positive constants, $\theta(t)$ is a nonrandom function, and $\widetilde B_1(t)$ and $\widetilde B_2(t)$ are correlated Brownian motions with $d\widetilde B_1(t) d\widetilde B_2(t)=\rho dt$ for some $\rho\in (-1,1)$. In this exercise, we discuss how to reduce this to the two-factor Vasicek model of Subsection 10.2.1, except that, instead of (10.2.6), the interest rate is given by (10.7.7), in which $\delta_0(t)$ is a nonrandom function of time.

\smallskip

\noindent (i) Define
\[
X(t) = \left[\begin{matrix} U(t) \\ R(t) \end{matrix}\right], \;
K = \left[\begin{matrix} \lambda_1 & 0 \\ -1 & \lambda_2 \end{matrix}\right], \;
\Sigma = \left[\begin{matrix} \sigma_1 & 0 \\ 0 & \sigma_2 \\ \end{matrix}\right]
\]
\[
\Theta(t) =\left[\begin{matrix} 0 \\ \theta(t) \end{matrix}\right], \;
\widetilde B(t) = \left[\begin{matrix} \widetilde B_1(t) \\ \widetilde B_2(t) \end{matrix}\right],
\]
so that (10.7.10) and (10.7.11) can be written in vector notation as
\[
dX(t) = \Theta(t) dt - KX(t) dt + \Sigma d\widetilde B(t). \tag{10.7.12}
\]
Now set
\[
\widehat X(t) = X(t) - e^{-Kt} \int_0^t e^{Ku}\Theta(u) du.
\]
Show that
\[
d\widehat X(t) = -K \widehat X(t) dt  + \Sigma d\widetilde B(t). \tag{10.7.13}
\]

\begin{proof}
\begin{eqnarray*}
d\widehat X(t) &=& dX(t)+Ke^{-Kt}\int_0^te^{Ku}\Theta(u)dudt-\Theta(t)dt\\
&=& -KX(t)dt+\Sigma d\widetilde B(t) +Ke^{-Kt}\int_0^te^{Ku}\Theta(u)dudt\\
&=& -K\widehat X(t)dt +\Sigma d\widetilde B(t).
\end{eqnarray*}
\end{proof}

\begin{remark}
The definition of $\widehat X$ is motivated by the observation that $Y$ has homogeneous $dt$ term in (10.2.4)-(10.2.5) and $e^{Kt}$ is the integrating factor for $X$: $d\left(e^{Kt}X(t)\right) = e^{Kt} \left(dX(t)-KX(t)dt\right)$.
\end{remark}

\noindent (ii) With
\[
C = \left[\begin{matrix} \frac{1}{\sigma_1} & 0 \\ -\frac{\rho}{\sigma_1\sqrt{1-\rho^2}} & \frac{1}{\sigma_2\sqrt{1-\rho^2}} \end{matrix}\right],
\]
define $Y(t) = C\widehat X(t)$, $\widetilde W(t) = C\Sigma\widetilde B(t)$. Show that the components of $\widetilde W_1(t)$ and $\widetilde W_2(t)$ are independent Brownian motions and
\[
dY(t) = -\Lambda Y(t) dt + d\widetilde W(t), \tag{10.7.14}
\]
where
\[
\Lambda = CKC^{-1} = \left[\begin{matrix} \lambda_1 & 0 \\ \frac{\rho\sigma_2(\lambda_2-\lambda_1)-\sigma_1}{\sigma_2\sqrt{1-\rho^2}} & \lambda_2\end{matrix}\right].
\]
Equation (10.7.14) is the vector form of the canonical two-factor Vasicek equations (10.2.4) and (10.2.5).

\begin{proof}
\[
\widetilde W(t)=C \Sigma \widetilde B(t)=\left(\begin{matrix} \frac{1}{\sigma_1} & 0 \\ -\frac{\rho}{\sigma_1\sqrt{1-\rho^2}} & \frac{1}{\sigma_2\sqrt{1-\rho^2}} \end{matrix}\right)
\left(\begin{matrix} \sigma_1 & 0 \\ 0 & \sigma_2 \end{matrix}\right) \widetilde B(t) = \left(\begin{matrix} 1 & 0 \\ -\frac{\rho}{\sqrt{1-\rho^2}} & \frac{1}{\sqrt{1-\rho^2}} \end{matrix}\right) \widetilde B(t).
\]
So $\widetilde W$ is a martingale with $\langle \widetilde W^1 \rangle(t) = \langle \widetilde B^1 \rangle(t)=t$,
\[
\langle \widetilde W^2 \rangle(t)=\left\langle -\frac{\rho}{\sqrt{1-\rho^2}}\widetilde B^1 + \frac{1}{\sqrt{1-\rho^2}}\widetilde B^2 \right\rangle(t)= \frac{\rho^2t}{1-\rho^2}+\frac{t}{1-\rho^2}-2\frac{\rho}{1-\rho^2}\rho t=\frac{\rho^2+1-2\rho^2}{1-\rho^2}t=t,
\]
and
\[
\langle \widetilde W^1,\widetilde W^2 \rangle(t) = \left\langle \widetilde B^1,-\frac{\rho}{\sqrt{1-\rho^2}}\widetilde B^1 +\frac{1}{\sqrt{1-\rho^2}}\widetilde B^2 \right\rangle(t)=-\frac{\rho t}{\sqrt{1-\rho^2}}+\frac{\rho t}{\sqrt{1-\rho^2}}=0.
\]
Therefore $\widetilde W$ is a two-dimensional BM. Moreover, $dY_t = Cd\widehat X_t = -CK\widehat X_t dt+C\Sigma d\widetilde B_t=-CKC^{-1}Y_tdt+d\widetilde W_t=-\Lambda Y_tdt+d\widetilde W_t$, where
\begin{eqnarray*}
\Lambda &=& CKC^{-1}= \left(\begin{matrix} \frac{1}{\sigma_1} & 0 \\ -\frac{\rho}{\sigma_1\sqrt{1-\rho^2}} & \frac{1}{\sigma_2\sqrt{1-\rho^2}} \end{matrix}\right)
\left(\begin{matrix} \lambda_1 & 0 \\ -1 & \lambda_2 \end{matrix}\right)
\cdot \frac{1}{|C|}
\left(\begin{matrix} \frac{1}{\sigma_2\sqrt{1-\rho^2}} & 0 \\ \frac{\rho}{\sigma_1\sqrt{1-\rho^2}} & \frac{1}{\sigma_1} \end{matrix}\right)\\
&=& \left(\begin{matrix} \frac{\lambda_1}{\sigma_1} & 0 \\ -\frac{\rho\lambda_1}{\sigma_1\sqrt{1-\rho^2}} -\frac{1}{\sigma_2\sqrt{1-\rho^2}}& \frac{\lambda_2}{\sigma_2\sqrt{1-\rho^2}} \end{matrix}\right)
\left(\begin{matrix} \sigma_1 & 0 \\ \rho\sigma_2  & \sigma_2\sqrt{1-\rho^2} \end{matrix}\right) \\
&=&\left(\begin{matrix} \lambda_1 & 0 \\ \frac{\rho\sigma_2(\lambda_2-\lambda_1)-\sigma_1}{\sigma_2\sqrt{1-\rho^2}} & \lambda_2 \end{matrix}\right).
\end{eqnarray*}
\end{proof}

\noindent (iii) Obtain a formula for $R(t)$ of the form (10.7.7). What are $\delta_0(t)$, $\delta_1$, and $\delta_2$?

\begin{solution}
\begin{eqnarray*}
X(t) &=& \widehat X(t) + e^{-Kt}\int_0^te^{Ku}\Theta(u)du = C^{-1}Y_t + e^{-Kt}\int_0^te^{Ku}\Theta(u)du \\
&=& \left(\begin{matrix} \sigma_1 & 0 \\ \rho\sigma_2 & \sigma_2\sqrt{1-\rho^2} \end{matrix}\right)
\left(\begin{matrix} Y_1(t) \\ Y_2(t) \end{matrix}\right) + e^{-Kt}\int_0^te^{Ku}\Theta(u)du \\
&=& \left(\begin{matrix} \sigma_1Y_1(t) \\ \rho\sigma_2Y_1(t)+ \sigma_2\sqrt{1-\rho^2}Y_2(t) \end{matrix}\right)+e^{-Kt}\int_0^te^{Ku}\Theta(u)du.
\end{eqnarray*}
So
\[
R(t)=X_2(t)=\rho\sigma_2Y_1(t)+\sigma_2\sqrt{1-\rho^2}Y_2(t)+\delta_0(t),
\]
where $\delta_0(t)$ is the second coordinate of $e^{-Kt}\int_0^te^{Ku}\Theta(u)du$ and can be derived explicitly by Lemma 10.2.3. Note
\begin{eqnarray*}
 e^{-Kt}\int_0^te^{Ku}\Theta(u)du
&=& \int_0^t \left[\begin{matrix} e^{\lambda_1(u-t)} & 0\\ * & e^{\lambda_2(u-t)} \end{matrix}\right] \left[\begin{matrix} 0 \\ \theta(u) \end{matrix}\right] du \\
&=& \int_0^t \left[\begin{matrix} 0 \\ \theta(u) e^{\lambda_2(u-t)}\end{matrix}\right] du
\end{eqnarray*}
Then $\delta_0(t) = e^{-\lambda_2t} \int_0^t e^{\lambda_2u}\theta(u) du$, $\delta_1=\rho\sigma_2$, and $\delta_2=\sigma_2\sqrt{1-\rho^2}$.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 10.5 (Correlation between long rate and short rate in the one-factor Vasicek model).} The one-factor Vasicek model is the one-factor Hull-White model of Example 6.5.1 with constant parameters,
\[
dR(t) = (a-bR(t))dt + \sigma d\widetilde W(t), \tag{10.7.15}
\]
where $a$, $b$, and $\sigma$ are positive constants and $\widetilde W(t)$ is a one-dimensional Brownian motion. In this model, the price at time $t\in [0,T]$ of the zero-coupon bond maturing at time $T$ is
\[
B(t,T) = e^{-C(t,T)R(t) - A(t,T)},
\]
where $C(t,T)$ and $A(t,T)$ are given by (6.5.10) and (6.5.11):
\begin{eqnarray*}
C(t,T) &=& \int_t^T e^{-\int_t^s bdv}ds = \frac{1}{b}\left(1 - e^{-b(T-t)}\right), \\
A(t,T) &=& \int_t^T \left(aC(s,t)-\frac{1}{2}\sigma^2C^2(s,T)\right)ds \\
&=& \frac{2ab-\sigma^2}{2b^2}(T-t) + \frac{\sigma^2-ab}{b^3}\left(1-e^{-b(T-t)}\right) - \frac{\sigma^2}{4b^3}\left(1-e^{-2b(T-t)}\right).
\end{eqnarray*}

In the spirit of the discussion of the short rate and the long rate in Subsection 10.2.1, we fix a positive relative maturity $\bar\tau$ and define the long rate $L(t)$ at time $t$ by (10.2.30):
\[
L(t) = -\frac{1}{\bar\tau} \log B(t,t+\bar\tau).
\]
Show that changes in $L(t)$ and $R(t)$ are perfectly correlated (i.e., for any $0\le t_1 < t_2$, the correlation coefficient between $L(t_2)-L(t_1)$ and $R(t_2)-R(t_1)$ is one). This characteristic of one-factor models caused the development of models with more than one factor.

\begin{proof} We note $C(t,T)$ and $A(t,T)$ are dependent only on $T-t$. So $C(t,t+\bar\tau)$ and $A(t,t+\bar \tau)$ are constants when $\bar\tau$ is fixed. So
\begin{eqnarray*}
\frac{d}{dt}L(t) &=& -\frac{B(t,t+\bar\tau)[-C(t,t+\bar\tau)R'(t)-A(t,t+\bar\tau)]}{\bar\tau B(t,t+\bar\tau)}\\
&=&\frac{1}{\bar\tau}[C(t,t+\bar\tau)R'(t)+A(t,t+\bar\tau)]\\
&=&\frac{1}{\bar\tau}[C(0,\bar\tau)R'(t)+A(0,\bar\tau)].
\end{eqnarray*}
Integrating from $t_1$ to $t_2$ on both sides, we have $L(t_2)-L(t_1)=\frac{1}{\bar\tau}C(0,\bar\tau)[R(t_2)-R(t_1)]+\frac{1}{\bar\tau}A(0,\bar\tau)(t_2-t_1)$. Since $L(t_2)-L(t_1)$ is a linear transformation of $R(t_2)-R(t_1)$, their correlation is 1.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 10.6 (Degenerate two-factor Vasicek model).} In the discussion of short rates and long rates in the two-factor Vasicek model of Subsection 10.2.1, we made the assumptions that $\delta_2 \ne 0$ and $(\lambda_1 - \lambda_2) \delta_1 + \lambda_{21} \delta_2 \ne 0$ (see Lemma 10.2.2). In this exercise, we show that if either of these conditions is violated, the two-factor Vasicek model reduces to a one-factor model, for which long rates and short rates are perfectly correlated (see Exercise 10.5).

\smallskip

\noindent (i) Show that if $\delta_2 = 0$ (and $\delta_0 > 0$, $\delta_1 > 0$), then the short rate $R(t)$ given by the system of equations (10.2.4)-(10.2.6) satisfies the one-dimensional stochastic differential equation
\[
dR(t) = (a-bR(t))dt + d\widetilde W_1(t). \tag{10.7.16}
\]
Define $a$ and $b$ in terms of the parameters in (10.2.4)-(10.2.6).

\begin{proof}
If $\delta_2=0$, then
\begin{eqnarray*}
dR(t)
&=& \delta_1 \left(-\lambda_1Y_1(t)dt+d\widetilde W_1(t)\right) \\
&=& \delta_1\left[\left(\frac{\delta_0}{\delta_1}-\frac{R(t)}{\delta_1}\right)\lambda_1dt+d\widetilde W_1(t)\right] \\
&=& (\delta_0\lambda_1-\lambda_1R(t))dt+\delta_1d\widetilde W_1(t).
\end{eqnarray*}
So $a=\delta_0\lambda_1$ and $b=\lambda_1$.
\end{proof}

\noindent (ii) Show that if $(\lambda_1 - \lambda_2) \delta_1 + \lambda_{21} \delta_2 = 0$ (and $\delta_0 > 0$, $\delta_1^2 + \delta_2^2 \ne 0$), then the short rate $R(t)$ given by the system of equations (10.2.4)-(10.2.6) satisfies the one-dimensional stochastic differential equation
\[
dR(t) = (a-bR(t))dt + \sigma d\widetilde B(t). \tag{10.7.17}
\]
Define $a$ and $b$ in terms of the parameters in (10.2.4)-(10.2.6) and define the Brownian motion $\widetilde B(t)$ in terms of the independent Brownian motions $\widetilde W_1(t)$ and $\widetilde W_2(t)$ in (10.2.4) and (10.2.5).

\begin{proof}
\begin{eqnarray*}
dR(t) &=& \delta_1dY_1(t)+\delta_2dY_2(t)\\
&=&-\delta_1\lambda_1Y_1(t)dt+\lambda_1d\widetilde W_1(t)-\delta_2\lambda_{21}Y_1(t)dt-\delta_2\lambda_2Y_2(t)dt+\delta_2d\widetilde W_2(t)\\
&=&-Y_1(t)(\delta_1\lambda_1+\delta_2\lambda_{21})dt-\delta_2\lambda_2Y_2(t)dt+\delta_1d\widetilde W_1(t)+\delta_2d\widetilde W_2(t)\\
&=&-Y_1(t)\lambda_2\delta_1dt-\delta_2\lambda_2Y_2(t)dt+\delta_1d\widetilde W_1(t)+\delta_2d\widetilde W_2(t)\\
&=&-\lambda_2(Y_1(t)\delta_1+Y_2(t)\delta_2)dt+\delta_1d\widetilde W_1(t)+\delta_2d\widetilde W_2(t)\\
&=&-\lambda_2(R_t-\delta_0)dt+\sqrt{\delta_1^2+\delta_2^2}\left[\frac{\delta_1}{\sqrt{\delta_1^2+\delta_2^2}}d\widetilde W_1(t)+\frac{\delta_2}{\sqrt{\delta_1^2+\delta_2^2}}d\widetilde W_2(t)\right].
\end{eqnarray*}
So $a=\lambda_2\delta_0$, $b=\lambda_2$, $\sigma=\sqrt{\delta_1^2+\delta_2^2}$ and $\widetilde B(t)=\frac{\delta_1}{\sqrt{\delta_1^2+\delta_2^2}}\widetilde W_1(t)+\frac{\delta_2}{\sqrt{\delta_1^2+\delta_2^2}}\widetilde W_2(t)$.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 10.7 (Forward measure in the two-factor Vasicek model).} Fix a maturity $T>0$. In the two-factor Vasicek model of Subsection 10.2.1, consider the $T$-forward measure $\widetilde {\mathbb P}^T$ of Definition 9.4.1:
\[
\widetilde {\mathbb P}^T(A) = \frac{1}{B(0,T)} \int_A D(T) d \widetilde {\mathbb P} \; \mbox{for all $A\in {\cal F}$.}
\]

\smallskip

\noindent (i) Show that the two-dimensional $\widetilde {\mathbb P}^T$-Brownian motion $\widetilde W_1^T(t)$, $\widetilde W_2^T(t)$ of (9.2.5) are
\[
\widetilde W_j^T(t) = \int_0^t C_j(T-u)du + \widetilde W_j(t), \; \mbox{$j=1,2$,} \tag{10.7.18}
\]
where $C_1(\tau)$ and $C_2(\tau)$ are given by (10.2.26)-(10.2.28).

\begin{proof} We use the canonical form of the model as in formulas (10.2.4)-(10.2.6). By (10.2.20),
\begin{eqnarray*}
dB(t,T) &=& df(t,Y_1(t),Y_2(t)) \\
&=& d e^{-Y_1(t)C_1(T-t)-Y_2(t)C_2(T-t)-A(T-t)}\\
&=& (\cdots) dt + B(t,T)[-C_1(T-t)d\widetilde W_1(t)-C_2(T-t)d\widetilde W_2(t)]\\
&=& (\cdots) dt + B(t,T)(-C_1(T-t),-C_2(T-t))\left(\begin{matrix}d\widetilde W_1(t) \\ d\widetilde W_2(t)\end{matrix}\right).
\end{eqnarray*}
So the volatility vector of $B(t,T)$ under $\widetilde {\mathbb P}$ is $(-C_1(T-t),-C_2(T-t))$. By (9.2.5), $\widetilde W_j^T(t)=\int_0^tC_j(T-u)du+\widetilde W_j(t)$ $(j=1,2)$ form a two-dimensional $\widetilde {\mathbb P}^T-$BM.
\end{proof}

\noindent (ii) Consider a call option on a bond maturing at time $\bar T>T$. The call expires at time $T$ and has strike price $K$. Show that at time zero the risk-neutral price of this option is
\[
B(0,T) \widetilde {\mathbb E}^T\left[\left(e^{-C_1(\bar T-T)Y_1(T)-C_2(\bar T-T)Y_2(T)-A(\bar T-T)} - K\right)^+\right]. \tag{10.7.19}
\]

\begin{proof}
Under the $T$-forward measure $\widetilde{\mathbb P}^T$, the numeraire is $B(t,T)$. By risk-neutral pricing, at time zero the risk-neutral price $V(0)$ of the option satisfies
\[
\frac{V(0)}{B(0,T)}=\widetilde {\mathbb E}^T\left[\frac{1}{B(T,T)} \left(e^{-C_1(\bar T-T)Y_1(T)-C_2(\bar T-T)Y_2(T)-A(\bar T-T)}-K \right)^+\right].
\]
Note $B(T,T)=1$, we get (10.7.19).
\end{proof}

\noindent (iii) Show that, under the $T$-forward measure $\widetilde {\mathbb P}^T$, the term
\[
X = -C_1(\bar T-T)Y_1(T) - C_2(\bar T-T)Y_2(T) -A(\bar T-T)
\]
appearing in the exponent in (10.7.19) is normally distributed.

\begin{proof}
Using (10.7.18), we can rewrite (10.2.4) and (10.2.5) as
\begin{eqnarray*}
\begin{cases}
dY_1(t)=-\lambda_1Y_1(t)dt+d\widetilde W_1^T(t)-C_1(T-t)dt\\
dY_2(t)=-\lambda_{21}Y_1(t)dt-\lambda_2Y_2(t)dt+d\widetilde W_2^T(t)-C_2(T-t)dt.
\end{cases}
\end{eqnarray*}
Then
\begin{eqnarray*}
\begin{cases}
Y_1(t)=Y_1(0)e^{-\lambda_1t}+\int_0^te^{\lambda_1(s-t)}d\widetilde W_1^T(s)-\int_0^tC_1(T-s)e^{\lambda_1(s-t)}ds\\
Y_2(t)=Y_0e^{-\lambda_2t}-\lambda_{21}\int_0^tY_1(s)e^{\lambda_2(s-t)}ds+\int_0^te^{\lambda_2(s-t)}d\widetilde W_2(s)-\int_0^tC_2(T-s)e^{\lambda_2(s-t)}ds.
\end{cases}
\end{eqnarray*}
Since $C_1$ is deterministic, $Y_1$ has Gaussian distribution. As the consequence, the second term in the expression of $Y_2$, $\int_0^tY_1(s)e^{\lambda_2(s-t)}ds$ also has Gaussian distribution and is uncorrelated to $\int_0^te^{\lambda_2(s-t)}d\widetilde W_2(s)$, since $\widetilde W_1^T$ and $\widetilde W_2^T$ are uncorrelated. Therefore, $(Y_1,Y_2)$ is jointly Gaussian and $X$ as a linear combination of them is also Gaussian.
\end{proof}

\noindent (iv) It is a straightforward but lengthy computation, like the computations in Exercise 10.1, to determine the mean and variance of the term $X$. Let us call its variance $\sigma^2$ and its mean $\mu-\frac{1}{2}\sigma^2$, so that we can write $X$ as
\[
X = \mu-\frac{1}{2}\sigma^2 - \sigma Z,
\]
where $Z$ is a standard normal random variable under $\widetilde {\mathbb P}^T$. Show that the call option price in (10.7.19) is
\[
B(0,T) \left(e^{\mu}N(d_+) - KN(d_-)\right),
\]
where
\[
d_{\pm} = \frac{1}{\sigma} \left(\mu-\log K \pm \frac{1}{2}\sigma^2\right).
\]

\begin{proof} The call option price in (10.7.19) is
\[
B(0,T) \widetilde{\mathbb E}^T\left[\left(e^X - K\right)^+\right]
\]
with $\widetilde{\mathbb E}^T[X] = \mu-\frac{1}{2}\sigma^2$ and $\mbox{Var}(X)=\sigma^2$. Comparing with the Black-Scholes formula for call options: if $dS_t=r S_tdt+\sigma S_td\widetilde W_t$, then
\[
\widetilde {\mathbb E}\left[e^{-r T}\left(S_0e^{\sigma W_T +(r-\frac{1}{2}\sigma^2)T}-K\right)^+\right]=S_0N(d_+)-Ke^{-r T}N(d_-)
\]
with $d_{\pm}=\frac{1}{\sigma\sqrt{T}}(\log \frac{S_0}{K}+(r\pm \frac{1}{2}\sigma^2)T)$, we can set in the Black-Scholes formula $r=\mu$, $T=1$ and $S_0=1$, then
\[
B(0,T) \widetilde{\mathbb E}^T\left[\left(e^X - K\right)^+\right]= B(0,T) e^{\mu}\widetilde{\mathbb E}^T\left[e^{-\mu}\left(e^X - K\right)^+\right]= B(0,T) \left(e^{\mu}N(d_+) - KN(d_-)\right)
\]
where $d_{\pm}=\frac{1}{\sigma}(-\log K+(\mu\pm\frac{1}{2}\sigma^2))$.
\end{proof}

\medskip

\noindent $\blacktriangleright$  {\bf Exercise 10.8 (Reversal of order of integration in forward rates).} The forward rate formula (10.3.5) with $v$ replacing $T$ states that
\[
f(t,v) = f(0,v) + \int_0^t \alpha(u,v)du + \int_0^t \sigma(u,v)dW(u).
\]
Therefore
\[
-\int_t^T f(t,v)dv = - \int_t^T \left[f(0,v)+\int_0^t\alpha(u,v)du + \int_0^t\sigma(u,v)dW(u)\right]dv. \tag{10.7.20}
\]

\smallskip

\noindent (i) Define 
\[
\widehat \alpha(u,t,T) = \int_t^T \alpha(u,v)dv, \; 
\widehat \sigma(u,t,T) = \int_t^T \sigma(u,v)dv.
\]
Show that if we reverse the order of integration in (10.7.20), we obtain the equation
\[
-\int_t^T f(t,v)dv = - \int_t^T f(0,v)dv - \int_0^t \widehat(u,t,T)du - \int_0^t \widehat\sigma(u,t,T)dW(u). \tag{10.7.21}
\]
(In one case, this is a reversal of the order of two Riemann integrals, a step that uses only the theory of ordinary calculus. In the other case, the order of a Riemann and an It\^{o} integral are being reversed. This step is justified in the appendix of [83]. You may assume without proof that this step is legitimate.)
\begin{proof} Starting from (10.7.20), we have
\begin{eqnarray*}
-\int_t^T f(t,v)dv 
&=& - \int_t^T f(0,v)dv - \int_{\{t\le v\le T, 0 \le u \le t\}} \alpha(u,v)dudv - \int_{\{t\le v\le T, 0 \le u \le t\}} \sigma(u,v)dW(u)dv \\
&=& -\int_t^T f(0,v)dv - \int_0^t du \int_t^T \alpha(u,v)dv - \int_0^t dW(u) \int_t^T \sigma(u,v)dv \\
&=& -\int_t^T f(0,v)dv - \int_0^t \widehat \alpha(u,t,T)du - \int_0^t \widehat \sigma(u,t,T) dW(u).
\end{eqnarray*}
\end{proof}

\noindent (ii) Take the differential with respect to $t$ in (10.7.21), remembering to get two terms from each of the integrals $\int_0^t \widehat\alpha(u,t,T)du$ and $\int_0^t \widehat\sigma(u,t,T)dW(u)$ because one must differentiate with respect to each of the two $t$s appearing in these integrals.
\begin{solution}
Differentiating both sides of (10.7.21), we have
\begin{eqnarray*}
& & d\left(-\int_t^T f(t,v)dv\right) \\
&=& f(0,t)dt - \widehat \alpha(t,t,T)dt - \int_0^t \frac{\partial}{\partial t} \widehat \alpha(u,t,T)du 
    - \widehat \sigma(t,t,T)dW(t) - \int_0^t \frac{\partial}{\partial t} \widehat\sigma(u,t,T) dW(u) \\
&=& f(0,t)dt - \int_t^T \alpha(t,v)dv dt + \int_0^t \alpha(u,t) dudt - \int_t^T \sigma(t,v)dvdW(t) + \int_0^t \sigma(u,t)dW(u)dt.
\end{eqnarray*}
\end{solution}

\noindent (iii) Check that your formula in (ii) agrees with (10.3.10).
\begin{proof} We note $R(t) = f(t,t) = f(0,t) + \int_0^t \alpha(u,v)du + \int_0^t \sigma(u,v) dW(u)$. From (ii), we therefore have
\[
d\left(-\int_t^T f(t,v)dv\right) = R(t) dt - \alpha^*(t,T) dt - \sigma^*(t,T)dW(t),
\]
which is (10.3.10).
\end{proof}


\medskip

\noindent $\blacktriangleright$  {\bf Exercise 10.9 (Multifactor HJM model).} Suppose the Heath-Jarrow-Morton model is driven by a $d$-dimensional Brownian motion, so that $\sigma(t,T)$ is also a $d$-dimensional vector and the forward rate dynamics are given by
\[
df(t,T) = \alpha(t,T)dt + \sum_{j=1}^d \sigma_j(t,T) dW_j(t).
\]

\smallskip

\noindent (i) Show that (10.3.16) becomes
\[
\alpha(t,T) = \sum_{j=1}^d \sigma_j(t,T) [\sigma_j^*(t,T) + \Theta_j(t)].
\]

\begin{proof} We first derive the SDE for the discounted bond price $D(t)B(t,T)$. We first note
\begin{eqnarray*}
d\left(-\int_t^T f(t,v)dv\right) &=& f(t,t)dt - \int_t^T df(t,v)dv = R(t)dt - \int_t^T \left[\alpha(t,v)dt+\sum_{j=1}^d \sigma_j(t,v) dW_j(t)\right]dv \\
&=& R(t) dt - \alpha^*(t,T)dt - \sum_{j=1}^d \sigma_j^*(t,T) dW_j(t).
\end{eqnarray*}
The applying It\^{o}'s formula, we have
\begin{eqnarray*}
dB(t,T) 
&=& d e^{-\int_t^T f(t,v)dv} = B(t,T)\left[d\left(-\int_t^T f(t,v)dv\right) + \frac{1}{2}\sum_{j=1}^d (\sigma_j^*(t,T))^2 dt \right] \\
&=& B(t,T) \left[R(t) - \alpha^*(t,T) + \frac{1}{2}\sum_{j=1}^d (\sigma_j^*(t,T))^2\right]dt - B(t,T) \sum_{j=1}^d \sigma_j^*(t,T) dW_j(t).
\end{eqnarray*}
Using integration-by-parts formula, we have
\begin{eqnarray*}
d(D(t)B(t,T)) &=& D(t)[-B(t,T)R(t)dt+dB(t,T)] \\
&=& D(t)B(t,T) \left[\left( - \alpha^*(t,T) + \frac{1}{2}\sum_{j=1}^d (\sigma_j^*(t,T))^2\right)dt - \sum_{j=1}^d \sigma_j^*(t,T) dW_j(t)\right]
\end{eqnarray*}

If no-arbitrage condition holds, we can find a risk-neutral measure $\widetilde {\mathbb P}$ under which
\[
\widetilde W(t) = \int_0^t\Theta(u)du + W(t)
\]
is a $d$-dimensional Brownian motion for some $d$-dimensional adapted process $\Theta(t)$. This implies
\[
- \alpha^*(t,T) + \frac{1}{2}\sum_{j=1}^d (\sigma_j^*(t,T))^2 + \sum_{j=1}^d \sigma_j^*(t,T)\Theta_j(t)= 0
\]
Differentiating both sides w.r.t. $T$, we obtain
\[
-\alpha(t,T) + \sum_{j=1}^d \sigma_j^*(t,T) \sigma_j(t,T) + \sum_{j=1}^d \sigma_j(t,T) \Theta_j(t) = 0.
\] 
Or equivalently,
\[
\alpha(t,T) = \sum_{j=1}^d \sigma_j(t,T) [\sigma_j^*(t,T)+\Theta_j(t)].
\]
\end{proof}

\noindent (ii) Suppose there is an adapted, $d$-dimensional process
\[
\Theta(t) = (\Theta_1(t), \cdots, \Theta_d(t))
\]
satisfying this equation for all $0 \le t \le T \le \overline{T}$. Show that if there are maturities $T_1$, $\cdots$, $T_d$ such that the $d\times d$ matrix $(\sigma_j(t,T_i))_{i,j}$ is nonsingular, then $\Theta(t)$ is unique.

\begin{proof}
Let $\sigma(t,T) = [\sigma_1(t,T), \cdots, \sigma_d(t,T)]^{tr}$ and $\sigma^*(t,T) = [\sigma_1^*(t,T), \cdots, \sigma_d^*(t,T)]^{tr}$. Then the no-arbitrage condition becomes
\[
\alpha(t,T) - (\sigma^*(t,T))^{tr} \sigma(t,T) = (\sigma(t,T))^{tr}\Theta(t).
\]
Let $T$ iterate $T_1$, $\cdots$, $T_d$, we have a matrix equation
\[
\left[\begin{matrix}
\alpha(t,T_1) - (\sigma^*(t,T_1))^{tr} \sigma(t,T_1)\\
\vdots \\
\alpha(t,T_d) - (\sigma^*(t,T_d))^{tr} \sigma(t,T_d)
\end{matrix}\right]
=
\left[\begin{matrix}
\sigma_1(t,T_1) & \ldots & \sigma_d(t,T_1) \\
\vdots & \ddots & \vdots \\
\sigma_1(t,T_d) & \ldots & \sigma_d(t,T_d)
\end{matrix}\right]
\left[\begin{matrix}
\Theta_1(t) \\
\vdots\\
\Theta_d(t) 
\end{matrix}\right]
\]
Therefore, if $(\sigma_j(t,T_i))_{i,j}$ is nonsingular, $\Theta(t)$ can be uniquely solved from the above matrix equation.
\end{proof}


\medskip

\noindent $\blacktriangleright$  {\bf Exercise 10.10.} (i) Use the ordinary differential equations (6.5.8) and (6.5.9) satisfied by the functions $A(t,T)$ and $C(t,T)$ in the one-factor Hull-White model to show that this model satisfies the HJM no-arbitrage condition (10.3.27)
\begin{proof}
Recall (6.5.8) and (6.5.9) are 
\[
\begin{cases}
C'(t,T) = b(t) C(t,T) -1  \\
A'(t,T) = -a(t)C(t,T)+\frac{1}{2}\sigma^2(t)C^2(t,T). 
\end{cases}
\]
Then by $\beta(t,r) = a(t)-b(t)r$ and $\gamma(t,r)=\sigma(t)$ (p. 430), we have
\begin{eqnarray*}
& & \frac{\partial}{\partial T} C(t,T) \beta(t,R(t)) + R(t)\frac{\partial}{\partial T} C'(t,T) + \frac{\partial}{\partial T} A'(t,T) \\
&=& \frac{\partial}{\partial T} C(t,T) \beta(t,R(t)) + R(t) b(t) \frac{\partial }{\partial T}C(t,T) + \left[-a(t)\frac{\partial}{\partial T}C(t,T)+\sigma^2(t)C(t,T)\frac{\partial}{\partial T}C(t,T)\right] \\
&=& \frac{\partial}{\partial T} C(t,T) \left[\beta(t,R(t))+ R(t) b(t)-a(t)+\sigma^2(t)C(t,T)\right] \\
&=& \left(\frac{\partial}{\partial T} C(t,T)\right)C(t,T)\gamma^2(t,R(t)).
\end{eqnarray*}
\end{proof}

\noindent (ii) Use the ordinary differential equations (6.5.14) and (6.5.15) satisfied by the functions $A(t,T)$ and $C(t,T)$ in the one-factor Cox-Ingersoll-Ross model to show that this model satisfies the HJM no-arbitrage condition (10.3.27).
\begin{proof}
Recall (6.5.14) and (6.5.15) are
\[
\begin{cases}
C'(t,T) = bC(t,T) + \frac{1}{2}\sigma^2C^2(t,T) - 1\\
A'(t,T) = -a C(t,T).
\end{cases}
\]
Then by $\beta(t,r) = a-br$ and $\gamma(t,r)=\sigma\sqrt{r}$ (p. 430), we have
\begin{eqnarray*}
& & \frac{\partial}{\partial T} C(t,T) \beta(t,R(t)) + R(t)\frac{\partial}{\partial T} C'(t,T) + \frac{\partial}{\partial T} A'(t,T) \\
&=& \frac{\partial}{\partial T} C(t,T) \beta(t,R(t)) + R(t) \left[b \frac{\partial}{\partial T} C(t,T)+\sigma^2C(t,T)\frac{\partial}{\partial T}C(t,T)\right] -a\frac{\partial}{\partial T}C(t,T)  \\
&=& \frac{\partial}{\partial T} C(t,T) \left[\beta(t,R(t))+ R(t) b+R(t)\sigma^2 C(t,T)-a\right] \\
&=& \left(\frac{\partial}{\partial T} C(t,T)\right)C(t,T)\gamma^2(t,R(t)).
\end{eqnarray*}
\end{proof}

\medskip

\noindent $\blacktriangleright$  {\bf Exercise 10.11.} Let $\delta > 0$ be given. Consider an interest rate swap paying a fixed interest rate $K$ and receiving backset LIBOR $L(T_{j-1}, T_{j-1})$ on a principal of 1 at each of the payment dates $T_j=\delta j$, $j=1, 2, \cdots, n+1$. Show that the value of the swap is 
\[
\delta K\sum_{j=1}^{n+1} B(0,T_j) - \delta \sum_{j=1}^{n+1} B(0, T_j) L(0, T_{j-1}).\tag{10.7.22}
\]
{\it Remark 10.7.1} The {\it swap rate} is defined to be the value of $K$ that makes the initial value of the swap equal to zero. Thus, the swap rate is
\[
K = \frac{\sum_{j=1}^{n+1} B(0,T_j)L(0,T_{j-1})}{\sum_{j=1}^{n+1} B(0,T_j)}. \tag{10.7.23}
\]

\begin{proof} On each payment date $T_j$, the payoff of this swap contract is $\delta(K-L(T_{j-1},T_{j-1}))$. Its no-arbitrage price at time 0 is $\delta(KB(0,T_j)-B(0,T_j)L(0,T_{j-1}))$ by Theorem 10.4. So the value of the swap is
\[
\sum_{j=1}^{n+1}\delta[KB(0,T_j)-B(0,T_j)L(0,T_{j-1})]=\delta K\sum_{j=1}^{n+1}B(0,T_j)-\delta \sum_{j=1}^{n+1}B(0,T_j)L(0,T_{j-1}).
\]
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 10.12.} In the proof of Theorem 10.4.1, we showed by an arbitrage argument that the value at time $0$ of a payment of backset LIBOR $L(T,T)$ at time $T+\delta$ is $B(0,T+\delta)L(0,T)$. The risk-neutral price of this payment, computed at time zero, is
\[
\widetilde{\mathbb E}[D(T+\delta)L(T,T)].
\]
Use the definitions
\begin{eqnarray*}
L(T,T) &=& \frac{1-B(T,T+\delta)}{\delta B(T,T+\delta)},\\
B(0,T+\delta) &=& \widetilde{\mathbb E}[D(T+\delta)],
\end{eqnarray*}
and the properties of conditional expectations to show that
\[
\widetilde{\mathbb E}[D(T+\delta)L(T,T)] = B(0,T+\delta)L(0,T).
\]

\begin{proof} Since $L(T,T)=\frac{1-B(T,T+\delta)}{\delta B(T,T+\delta)} \in {\cal F}_T$, we have
\begin{eqnarray*}
\widetilde {\mathbb E}[D(T+\delta)L(T,T)] &=& \widetilde {\mathbb E}[\widetilde {\mathbb E}[D(T+\delta)L(T,T)|{\cal F}_T]]\\
&=&\widetilde {\mathbb E}\left[ \frac{1-B(T,T+\delta)}{\delta B(T,T+\delta)}\widetilde {\mathbb E}[D(T+\delta)|{\cal F}_T]\right] \\
&=&\widetilde {\mathbb E}\left[\frac{1-B(T,T+\delta)}{\delta B(T,T+\delta)}D(T)B(T,T+\delta)\right]\\
&=&\widetilde {\mathbb E}\left[\frac{D(T)-D(T)B(T,T+\delta)}{\delta}\right]\\
&=&\frac{B(0,T)-B(0,T+\delta)}{\delta}\\
&=& B(0,T+\delta)L(0,T).
\end{eqnarray*}
\end{proof}

\begin{remark}
An alternative proof is to use the $(T+\delta)$-forward measure $\widetilde{\mathbb P}^{T+\delta}$: the time-$t$ no-arbitrage price of a payoff $\delta L(T,T)$ at time $(T+\delta)$ is
\[
B(t,T+\delta) \widetilde {\mathbb E}^{T+\delta}[\delta L(T,T)] = B(t,T+\delta) \cdot \delta L(t,T),
\]
where we have used the observation that the forward rate process $L(t,T) = \frac{1}{\delta}\left(\frac{B(t,T)}{B(t,T+\delta)} - 1\right) \; (0 \le t \le T)$ is a martingale under the $(T+\delta)$-forward measure.
\end{remark}

\medskip

\section{Introduction to Jump Processes}

$\bigstar$ {\bf Comments}:

\medskip

1) A mathematically rigorous presentation of semimartingale theory and stochastic calculus can be found in He et al. \cite{HWY92} and Kallenberg \cite{Kallenberg02}.

\medskip

2) {\it Girsanov's theorem}. The most general version of Girsanov's theorem for local martingales can be found in He et al. \cite[page~340]{HWY92} (Theorem 12.13) or Kallenberg \cite[page~523]{Kallenberg02} (Theorem 26.9). It has the following form
\begin{theorem}[Girsanov's theorem for local martingales]
Let $Q=Z_t \cdot P$ on ${\cal F}_t$ for all $t\ge 0$, and consider a local $P$-martingale $M$ such that the process $[M,Z]$ has locally integrable variation and $P$-compensator $\langle M, Z \rangle$. Then $\bar M = M - Z^{-1}_- \cdot \langle M, Z \rangle$ is a local $Q$-martingale.\footnote{For the difference between the {\it predictable quadratic variation} $\langle \cdot \rangle$ (or {\it angle bracket process}) and the {\it quadratic variation} (or {\it square bracket process}), see He et al. \cite[page~185-187]{HWY92}. Basically, we have $[M, N]_t = M_0N_0 + \langle M^c, N^c \rangle_t + \sum_{s\le t}\Delta M_s \Delta N_s$ and $\langle M, N \rangle$ is the dual predictable projection of $[M, N]$ (i.e. $[M,N]_t- \langle M, N \rangle_t$ is a martingale).}
\end{theorem}
Applying Girsanov's theorem to Lemma 11.6.1, in order to change the intensity of a Poisson process via change of measure, we need to find a ${\mathbb P}$-martingale $L$ such that the Radon-Nikod\'{y}m derivative $Z_{\cdot} = d\widetilde {\mathbb P}/d{\mathbb P}$ satisfies the SDE
\[
dZ(t) = Z(t-) dL(t)
\]
and
\[
N(t)-\lambda t - \langle N(t)-\lambda t, L(t)\rangle = N(t) - \widetilde \lambda t
\]
is a martingale under $\widetilde {\mathbb P}$. This implies we should find $L$ such that
\[
 \langle N(t)-\lambda t, L(t)\rangle = (\widetilde\lambda -  \lambda) t.
\]
Assuming martingale representation theorem, we suppose $L(t) = L(0)+\int_0^t H(s) d(N(s)-\lambda s)$, then
\[
[N(t)-\lambda t, L(t)] = \int_0^t H(s) d [N(s)-\lambda s, N(s)-\lambda s] = \sum_{0<s\le t}H(s)(\Delta N(s))^2 = \int_0^t H(s) dN(s).
\]
So $\langle N(t)-\lambda t, L(t)\rangle = \int_0^t H(s) \lambda ds$. Solving the equation
\[
\int_0^t H(s) \lambda ds=(\widetilde\lambda -  \lambda) t,
\]
we get $H(s) = \frac{\widetilde\lambda -  \lambda}{\lambda}$. Combined, we conclude $Z$ should be determined by the equation (11.6.2)
\[
dZ(t) = \frac{\widetilde\lambda -  \lambda}{\lambda} Z(t-)  dM(t).
\]

\bigskip

\noindent $\blacktriangleright$ {\bf Exercise 11.1.} Let $M(t)$ be the compensated Poisson process of Theorem 11.2.4.

\smallskip

\noindent (i) Show that $M^2(t)$ is a submartingale.

\begin{proof} First, $M^2(t)=N^2(t)-2\lambda
tN(t) +\lambda^2t^2$. So ${\mathbb E}[M^2(t)]<\infty$. $\varphi(x)=x^2$ is a convex
function, by the conditional Jensen's inequality (Shreve \cite[page~70]{Shreve04b}, Theorem 2.3.2 (v)),
\[
{\mathbb E}[\varphi(M(t))|{\cal F}(s)]\ge \varphi({\mathbb E}[M(t)|{\cal F}(s)])=\varphi(M(s)), \;\forall s \le
t.
\]
So $M^2(t)$ is a submartingale.
\end{proof}

\noindent (ii) Show that $M^2(t)-\lambda t$ is a martingale.

\begin{proof} We note $M(t)=N(t)-\lambda t$ has independent and stationary
increment. So $\forall s\le t$,
\begin{eqnarray*}
& & {\mathbb E}[M^2(t)-M^2(s)|{\cal F}(s)] \\
&=& {\mathbb E}[(M(t)-M(s))^2|{\cal F}(s)]+{\mathbb E}[(M(t)-M(s))\cdot 2M(s)|{\cal F}(s)] \\
&=& {\mathbb E}[M^2(t-s)]+2M(s){\mathbb E}[M(t-s)] \\
&=& \mbox{Var}(N(t-s))+0 \\
&=& \lambda(t-s).
\end{eqnarray*}
That is, ${\mathbb E}[M^2(t)-\lambda t|{\cal F}(s)]=M^2(s)-\lambda s$.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 11.2.} Suppose we have observed a Poisson process up to time $s$, have seen that $N(s)=k$, and are interested in the values of $N(s+t)$ for small positive $t$. Show that
\begin{eqnarray*}
{\mathbb P}\{N(s+t)=k | N(s)=k\} = 1 - \lambda t + O(t^2), \\
{\mathbb P}\{N(s+t)=k+1 | N(s) = k\} = \lambda t + O(t^2), \\
{\mathbb P}\{N(s+t)\ge k+2 | N(s)=k \} = O(t^2),
\end{eqnarray*}
where $O(t^2)$ is used to denote terms involving $t^2$ and higher powers of $t$.

\begin{proof} We note
\[
{\mathbb P}(N(s+t)=k|N(s)=k)={\mathbb P}(N(s+t)-N(s)=0|N(s)=k)={\mathbb P}(N(t)=0)=e^{-\lambda
t}=1-\lambda t+O(t^2).
\]
Similarly, we have
\[
{\mathbb P}(N(s+t)=k+1|N(s)=k)={\mathbb P}(N(t)=1)=\frac{(\lambda t)^1}{1!}e^{-\lambda
t}=\lambda t(1-\lambda t+O(t^2))=\lambda t+O(t^2),
\]
and
\[
{\mathbb P}(N(s+t) \ge k+2|N(t)=k)={\mathbb P}(N(t) \ge 2)=\sum_{k=2}^{\infty}\frac{(\lambda t)^k}{k!}e^{-\lambda t}=O(t^2).
\]
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 11.3 (Geometric Poisson process).} Let $N(t)$ be a Poisson process with intensity $\lambda > 0$, and let $S(0) > 0$ and $\sigma > -1$ be given. Using Theorem 11.2.3 rather than the It\^{o}-Doeblin formula for jump processes, show that
\[
S(t) = \exp \left\{ N(t)\log (\sigma+1) - \lambda \sigma t\right\} = (\sigma+1)^{N(t)} e^{-\lambda\sigma t}
\]
is a martingale.

\begin{proof}For any $t\le u$, we have
\begin{eqnarray*}
{\mathbb E} \left[\left.\frac{S(u)}{S(t)}\right|{\cal F}(t)\right]
&=&{\mathbb E} \left[(\sigma+1)^{N(t)-N(u)}e^{-\lambda\sigma(t-u)}|{\cal F}(t) \right] \\
&=& e^{-\lambda\sigma(t-u)}{\mathbb E}\left[(\sigma+1)^{N(t-u)}\right]\\
&=& e^{-\lambda\sigma(t-u)}{\mathbb E}[e^{N(t-u)\log(\sigma+1)}]\\
&=& e^{-\lambda\sigma(t-u)}\exp\left\{\lambda(t-u)\left(e^{\log(\sigma+1)}-1\right)\right\} \;\;\mbox{(by (11.3.4))}\\
&=& e^{-\lambda\sigma(t-u)}e^{\lambda\sigma(t-u)}\\
&=&1.
\end{eqnarray*}
So $S(t)={\mathbb E}[S(u)|{\cal F}(t)]$ and $S$ is a martingale.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 11.4.} Suppose $N_1(t)$ and $N_2(t)$ are Poisson processes with intensities $\lambda_1$ and $\lambda_2$, respectively, both defined on the same probability space $(\Omega, {\cal F}, {\mathbb P})$ and relative to the same filtration ${\cal F}(t)$, $t\ge 0$. Show that almost surely $N_1(t)$ and $N_2(t)$ can have no simultaneous jump. (Hint: Define the compensated Poisson processes $M_1(t) = N_1(t)-\lambda t$ and $M_2(t) = N_2(t) - \lambda_2 t$, which like $N_1$ and $N_2$ are independent. Use It\^{o} product rule for jump processes to compute $M_1(t)M_2(t)$ and take expectations.)

\begin{proof}The problem is ambiguous in that the
relation between $N_1$ and $N_2$ is not clearly stated. According to
page 524, paragraph 2, we would guess the condition should be that
$N_1$ and $N_2$ are independent.

Suppose $N_1$ and $N_2$ are independent. Define
$M_1(t)=N_1(t)-\lambda_1t$ and $M_2(t)=N_2(t)-\lambda_2t$. Then by
independence ${\mathbb E}[M_1(t)M_2(t)]={\mathbb E}[M_1(t)]{\mathbb E}[M_2(t)]=0$. Meanwhile, by
It\^{o}'s product formula (Corollary 11.5.5),
\[
M_1(t)M_2(t)=\int_0^tM_1(s-)dM_2(s)+\int_0^tM_2(s-)dM_1(s)+[M_1,M_2](t).
\]
Both $\int_0^tM_1(s-)dM_2(s)$ and $\int_0^tM_2(s-)dM_1(s)$ are
martingales. So taking expectation on both sides, we get
\[
0=0+{\mathbb E}\{[M_1,M_2](t)\}={\mathbb E}\left[\sum_{0<s\le t}\Delta N_1(s) \Delta
N_2(s)\right].
\]
Since $\sum_{0<s\le t}\Delta N_1(s) \Delta N_2(s)\ge 0$
a.s., we conclude $\sum_{0<s\le t}\Delta N_1(s) \Delta N_2(s)=0$
a.s. By letting $t=1,2,\cdots$, we can find a set $\Omega_0$ of
probability 1, so that $\forall \omega\in \Omega_0$, $\sum_{0<s\le
t}\Delta N_1(\omega; s)\Delta N_2(\omega; s)=0$ for all $t>0$. Therefore $N_1$ and
$N_2$ can have no simultaneous jump.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 11.5.} Suppose $N_1(t)$ and $N_2(t)$ are Poisson processes defined on the same probability space $(\Omega, {\cal F}, {\mathbb P})$ relative to the same filtration ${\cal F}(t)$, $t\ge 0$. Assume that almost surely $N_1(t)$ and $N_2(t)$ have no simultaneous jump. Show that, for each fixed $t$, the random variable $N_1(t)$ and $N_2(t)$ are independent. (Hint: Adapt the proof of Corollary 11.5.3.) (In fact, the whole path of $N_1$ is independent of the whole path of $N_2$, although you are not being asked to prove this stronger statement.)

\begin{proof} We shall prove the whole path of $N_1$
is independent of the whole path of $N_2$, following the scheme
suggested by page 489, paragraph 1.

Fix $s\ge 0$, we consider $X_t=u_1(N_1(t)-N_1(s))+u_2(N_2(t)-N_2(s))-\lambda_1(e^{u_1}-1)(t-s)-\lambda_2(e^{u_2}-1)(t-s)$,
$t>s$. Then by It\^{o}'s formula for jump process, we have
\begin{eqnarray*}
e^{X_t}-e^{X_s}&=&\int_s^te^{X_u}dX_u^c+\frac{1}{2}\int_s^te^{X_u}dX_u^cdX_u^c+\sum_{s<u\le
t}(e^{X_u}-e^{X_{u-}})\\
&=&\int_s^te^{X_u}[-\lambda_1(e^{u_1}-1)-\lambda_2(e^{u_2}-1)]du+\sum_{0<u\le
t}(e^{X_u}-e^{X_{u-}}).
\end{eqnarray*}
Since $\Delta X_t=u_1\Delta N_1(t)+u_2\Delta N_2(t)$ and $N_1$, $N_2$ have no simultaneous jump,
\[
e^{X_u}-e^{X_{u-}}=e^{X_{u-}}(e^{\Delta X_u}-1)=e^{X_{u-}}[(e^{u_1}-1)\Delta N_1(u)+(e^{u_2}-1)\Delta
N_2(u)].
\]
Note $\int_s^te^{X_{u}}du = \int_s^te^{X_{u-}}du$\footnote{On the interval $[s,t]$, the sample path $X_{\cdot}(\omega)$ has only finitely many jumps for each $\omega$, so the Riemann integrals of $X_{\cdot}$ and $X_{\cdot-}$ should agree with each other.}, we have
\begin{eqnarray*}
&
&e^{X_t}-e^{X_s}\\
&=&\int_s^te^{X_{u-}}[-\lambda_1(e^{u_1}-1)-\lambda_2(e^{u_2}-1)]du+\sum_{s<u\le
t}e^{X_{u-}}[(e^{u_1}-1)\Delta N_1(u)+(e^{u_2}-1)\Delta N_2(u)]\\
&=&\int_s^te^{X_{u-}} \left[(e^{u_1}-1)d(N_1(u)-\lambda_1u)-(e^{u_2}-1)d(N_2(u)-\lambda_2u)\right].
\end{eqnarray*}
Therefore, ${\mathbb E}[e^{X_t}]={\mathbb E}[e^{X_s}]=1$, which implies
\begin{eqnarray*}
{\mathbb E} \left[e^{u_1(N_1(t)-N_1(s))+u_2(N_2(t)-N_2(s))}\right]
&=& e^{\lambda_1(e^{u_1}-1)(t-s)}e^{\lambda_2(e^{u_2}-1)(t-s)} \\
&=& {\mathbb E}\left[e^{u_1(N_1(t)-N_1(s))}\right] {\mathbb E}\left[e^{u_2(N_2(t)-N_2(s))}\right].
\end{eqnarray*}
This shows $N_1(t)-N_1(s)$ is independent of $N_2(t)-N_2(s)$.

Now, suppose we have $0\le t_1<t_2<t_3<\cdots<t_n$, then the vector
$(N_1(t_1),\cdots,N_1(t_n))$ is independent of
$(N_2(t_1),\cdots,N_2(t_n))$ if and only if
$(N_1(t_1),N_1(t_2)-N_1(t_1),\cdots,N_1(t_n)-N_1(t_{n-1}))$ is
independent of
$(N_2(t_1),N_2(t_2)-N_2(t_1),\cdots,N_2(t_n)-N_2(t_{n-1}))$. Let
$t_0=0$, then
\begin{eqnarray*}
& &{\mathbb E} \left[e^{\sum_{i=1}^nu_i(N_1(t_i)-N_1(t_{i-1}))+\sum_{j=1}^nv_j(N_2(t_j)-N_2(t_{j-1}))} \right]\\
&=&{\mathbb E} \left[e^{\sum_{i=1}^{n-1}u_i(N_1(t_i)-N_1(t_{i-1}))+\sum_{j=1}^{n-1}v_j(N_2(t_j)-N_2(t_{j-1}))} \cdot \right. \\
& & \left. {\mathbb E}\left[\left.e^{u_n(N_1(t_n)-N_1(t_{n-1}))+v_n(N_2(t_n)-N_2(t_{n-1}))}\right|{\cal
F}(t_{n-1})\right]\right]\\
&=&{\mathbb E} \left[e^{\sum_{i=1}^{n-1}u_i(N_1(t_i)-N_1(t_{i-1}))+\sum_{j=1}^{n-1}v_j(N_2(t_j)-N_2(t_{j-1}))}\right]{\mathbb E}\left[e^{u_n(N_1(t_n)-N_1(t_{n-1}))+v_n(N_2(t_n)-N_2(t_{n-1}))}\right]\\
&=&{\mathbb E}\left[e^{\sum_{i=1}^{n-1}u_i(N_1(t_i)-N_1(t_{i-1}))+\sum_{j=1}^{n-1}v_j(N_2(t_j)-N_2(t_{j-1}))}\right]{\mathbb E}\left[e^{u_n(N_1(t_n)-N_1(t_{n-1}))}\right] {\mathbb E}\left[e^{v_n(N_2(t_n)-N_2(t_{n-1}))}\right],
\end{eqnarray*}
where the second equality comes from the independence of
$N_i(t_n)-N_i(t_{n-1})$ $(i=1,2)$ relative to ${\cal F}(t_{n-1})$
and the third equality comes from the result obtained just before. Working by induction, we have \begin{eqnarray*} &
&{\mathbb E}\left[e^{\sum_{i=1}^nu_i(N_1(t_i)-N_1(t_{i-1}))+\sum_{j=1}^nv_j(N_2(t_j)-N_2(t_{j-1}))}\right]\\
&=&\prod_{i=1}^n{\mathbb E}\left[e^{u_i(N_1(t_i)-N_1(t_{i-1}))}\right] \cdot \prod_{j=1}^n{\mathbb E}\left[e^{v_j(N_2(t_j)-N_2(t_{j-1}))}\right]\\
&=&{\mathbb E}\left[e^{\sum_{i=1}^nu_i(N_1(t_i)-N_1(t_{i-1}))}\right] \cdot {\mathbb E}\left[e^{\sum_{j=1}^nv_j(N_2(t_j)-N_2(t_{j-1}))}\right].
\end{eqnarray*}
This shows the whole path of $N_1$ is independent of the whole path
of $N_2$.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 11.6.} Let $W(t)$ be a Brownian motion and let $Q(t)$ be a compound Poisson process, both defined on the same probability space $(\Omega, {\cal F}, {\mathbb P})$ and relative to the same filtration ${\cal F}(t)$, $t\ge 0$. Show that, for each $t$, the random variables $W(t)$ and $Q(t)$ are independent. (In fact, the whole path of $W$ is independent of the whole path of $Q$, although you are not being asked to prove this stronger statement.)

\begin{proof}Let
$X_t = u_1W(t)-\frac{1}{2}u_1^2t+u_2Q(t)-\lambda t(\varphi(u_2)-1)$
where $\varphi$ is the moment generating function of the jump size
$Y$. It\^{o}'s formula for jump process yields
\begin{eqnarray*}
e^{X_t}-1=\int_0^te^{X_s}\left(u_1dW(s) - \frac{1}{2}u_1^2ds-\lambda
(\varphi(u_2)-1)ds\right)+\frac{1}{2}\int_0^te^{X_s}u_1^2ds+\sum_{0< s\le
t}(e^{X_s}-e^{X_{s-}}).
\end{eqnarray*}
Note $\Delta X_t=u_2\Delta Q(t)=u_2Y_{N(t)}\Delta N(t)$, where $N(t)$ is
the Poisson process associated with $Q(t)$. So
\[
e^{X_t}-e^{X_{t-}}=e^{X_{t-}}(e^{\Delta
X_t}-1)=e^{X_{t-}}(e^{u_2Y_{N(t)}}-1)\Delta N(t).
\]
Consider the
compound Poisson process $H_t=\sum_{i=1}^{N(t)}(e^{u_2Y_i}-1)$, then
\[
H_t-\lambda {\mathbb E}\left[e^{u_2Y_1}-1\right]t = H_t-\lambda (\varphi(u_2)-1)t
\]
is a martingale, $e^{X_t}-e^{X_{t-}}=e^{X_{t-}}\Delta H_t$ and
\begin{eqnarray*}
e^{X_t}-1
&=&\int_0^te^{X_s} \left(u_1dW(s)-\frac{1}{2}u_1^2ds-\lambda (\varphi(u_2)-1)ds \right)+\frac{1}{2}\int_0^te^{X_s}u_1^2ds+\int_0^te^{X_{s-}}dH_s\\
&=&\int_0^te^{X_s}u_1dW(s)+\int_0^te^{X_{s-}}d(H_s-\lambda
(\varphi(u_2)-1)s).
\end{eqnarray*}
This shows $e^{X_t}$ is a martingale and ${\mathbb E}\left[e^{X_t}\right]\equiv 1$. So
\[
{\mathbb E}\left[e^{u_1W(t)+u_2Q(t)}\right] = e^{\frac{1}{2}u_1t}e^{\lambda
t(\varphi(u_2)-1)t}={\mathbb E} \left[e^{u_1W(t)}\right] {\mathbb E}\left[e^{u_2Q(t)}\right].
\]
This shows $W(t)$ and $Q(t)$ are independent.
\end{proof}

\begin{remark}
It is easy to see that, if we follow the steps of solution to Exercise 11.5, firstly proving the independence of $W(t)-W(s)$ and $Q(t)-Q(s)$, then proving the independence of $(W(t_1),W(t_2)-W(t_1),\cdots, W(t_n)-W(t_{n-1}))$ and $(Q(t_1),Q(t_2)-Q(t_1),\cdots,Q(t_n)-Q(t_{n-1}))$ ($0 \le t_1 < t_2 < \cdots < t_n$), then we can show the whole path of $W$ is independent of the whole path of $Q$.
\end{remark}
\medskip

\noindent $\blacktriangleright$ {\bf Exercise 11.7.} Use Theorem 11.3.2 to prove that a compound Poisson process is Markov. In other words, show that, whenever we are given two times $0\le t \le T$ and a function $h(x)$, there is another function $g(t,x)$ such that
\[
{\mathbb E} [h(Q(T)) | {\cal F}(t) ] = g(t, Q(t)).
\]

\begin{proof} By Independence Lemma, we have
\[
{\mathbb E}[h(Q(T))|{\cal F}(t)]
= {\mathbb E}[h(Q(T)-Q(t)+Q(t))|{\cal F}(t)]
= {\mathbb E}[h(Q(T-t)+x)]|_{x=Q(t)}=g(t,Q(t)),
\]
where $g(t,x)={\mathbb E}[h(Q(T-t)+x)]$.
\end{proof}

\begin{thebibliography}{99}

\bibitem{AC92} Vladimir I. Arnold and Roger Cooke. {\it Ordinary differential equations}, 3rd edition, Springer, 1992.

\bibitem{DS06} Freddy Delbaen and Walter Schachermayer. {\it The mathematics of arbitrage}. Springer, 2006.

\bibitem{DL91} 丁同仁、李承治：《常微分方程教程》，北京：高等教育出版社，1991.

\bibitem{Durrett10} Richard Durrett. {\it Probability: Theory and examples}, 4th edition. Cambridge University Press, New York, 2010.

\bibitem{HWY92} Sheng-wu He, Jia-gang Wang, and Jia-an Yan. {\it Semimartingale theory and stochastic calculus}. Science Press \& CRC Press Inc, 1992.

\bibitem{Hull00} John Hull. {\it Options, futures, and other derivatives}, 4th edition. Prentice-Hall International Inc., New Jersey, 2000.

\bibitem{Kallenberg02} Olav Kallenberg. {\it Foundations of modern probability}, 2nd edition. Springer, 2002.

\bibitem{Logan10} J. David Logan. {\it A first course in differential equations}, 2nd edition. Springer, New York, 2010.

\bibitem{Oksendal03} B. \O ksendal. {\it Stochastic differential equations: An introduction with applications}, 6th edition. Springer-Verlag, Berlin, 2003.

\bibitem{QG97} 钱敏平、龚光鲁：《随机过程论（第二版）》，北京：北京大学出版社，1997.10。

\bibitem{RY98} Daniel Revuz and Marc Yor. {\it Continous martingales and Brownian motion}, 3rd edition. Springer-Verlag, Berlin, 1998.

\bibitem{Shiryaev95} A. N. Shiryaev. {\it Probability}, 2nd edition. Springer, 1995.

\bibitem{Shiryaev99} A. N. Shiryaev. {\it Essentials of stochastic finance: facts, models, theory}. World Scientific, Singapore, 1999.

%\bibitem{Shreve04a} Steven Shreve. {\it Stochastic calculus for finance I: The binomial asset pricing model}. Springer-Verlag, New York, 2004.

\bibitem{Shreve04b} Steven Shreve. {\it Stochastic calculus for finance II: Continuous-time models}. Springer-Verlag, New York, 2004.

\bibitem{Wilmott95} Paul Wilmott. {\it The mathematics of financial derivatives: A student introduction}. Cambridge University Press, 1995.
\end{thebibliography}

\end{document}

% Version 1.0.8, 2015-03-13: completed Chapter 10.
% Version 1.0.7, 2015-01-23: added problems; revised comments/annotation.
% Version 1.0,   2007-08-20: first draft.



